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## 5.2: Vertex Form

- Last updated
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- Page ID 19707

- David Arnold
- College of the Redwoods

In the previous section, you learned that it is a simple task to sketch the graph of a quadratic function if it is presented in vertex form

\[f(x)=a(x-h)^{2}+k \nonumber \]

The goal of the current section is to start with the most general form of the quadratic function, namely

\[f(x)=a x^{2}+b x+c \nonumber \]

and manipulate the equation into vertex form. Once you have your quadratic function in vertex form, the technique of the previous section should allow you to construct the graph of the quadratic function. However, before we turn our attention to the task of converting the general quadratic into vertex form, we need to review the necessary algebraic fundamentals. Let’s begin with a review of an important algebraic shortcut called squaring a binomial.

## Squaring a Binomial

A monomial is a single algebraic term, usually constructed as a product of a number (called a coefficient) and one or more variables raised to nonnegative integral powers, such as \(-3 x^{2}\) or 14\(y^{3} z^{5}\). The key phrase here is “single term.” A binomial is an algebraic sum or difference of two monomials (or terms), such as \(x+2 y\) or \(3 a b^{2}-2 c^{3}\). The key phrase here is “two terms.”

To “square a binomial,” start with an arbitrary binomial, such as a+b, then multiply it by itself to produce its square (a + b)(a + b), or, more compactly, \((a+b)^{2}\). We can use the distributive property to expand the square of the binomial a + b.

\[\begin{aligned}(a+b)^{2} &=(a+b)(a+b) \\ &=a(a+b)+b(a+b) \\ &=a^{2}+a b+b a+b^{2} \end{aligned} \nonumber \]

Because ab = ba, we can add the two middle terms to arrive at the following property.

The square of the binomial a + b is expanded as follows.

\[(a+b)^{2}=a^{2}+2 a b+b^{2} \nonumber \]

## Example \(\PageIndex{1}\)

Expand \((x+4)^{2}\)

We could proceed as follows.

\[\begin{align*}(x+4)^{2} &=(x+4)(x+4) \\ &=x(x+4)+4(x+4) \\ &=x^{2}+4 x+4 x+16 \\ &=x^{2}+8 x+16 \end{align*} \]

Although correct, this technique will not help us with our upcoming task. What we need to do is follow the algorithm suggested by Property 3.

## Algorithm for Squaring a Binomial

To square the binomial a + b, proceed as follows:

- Square the first term to get \(a^2\).
- Multiply the first and second terms together, and then multiply the result by two to get 2ab.
- Square the second term to get \(b^2\).

Thus, to expand \((x + 4)^2\), we should proceed as follows.

- Square the first term to get \(x^2\)
- Multiply the first and second terms together and then multiply by two to get 8x.
- Square the second term to get 16.

Proceeding in this manner allows us to perform the expansion mentally and simply write down the solution.

\[(x+4)^{2}=x^{2}+2(x)(4)+4^{2}=x^{2}+8 x+16 \nonumber \]

Here are a few more examples. In each, we’ve written an extra step to help clarify the procedure. In practice, you should simply write down the solution without any intermediate steps.

\[\begin{array}{l}{(x+3)^{2}=x^{2}+2(x)(3)+3^{2}=x^{2}+6 x+9} \\ {(x-5)^{2}=x^{2}+2(x)(-5)+(-5)^{2}=x^{2}-10 x+25} \\ {\left(x-\frac{1}{2}\right)^{2}=x^{2}+2(x)\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)^{2}=x^{2}-x+\frac{1}{4}}\end{array} \nonumber \]

It is imperative that you master this shortcut before moving on to the rest of the material in this section.

## Perfect Square Trinomials

Once you’ve mastered squaring a binomial, as in

it’s a simple matter to identify and factor trinomials (three terms) having the form \(a^{2}+2 a b+b^{2}\). You simply “undo” the multiplication. Whenever you spot a trinomial whose first and third terms are perfect squares, you should suspect that it factors as follows.

\[a^{2}+2 a b+b^{2}=(a+b)^{2} \nonumber \]

A trinomial that factors according to this rule or pattern is called a perfect square trinomial.

For example, the first and last terms of the following trinomial are perfect squares.

\[x^{2}+16 x+64 \nonumber \]

The square roots of the first and last terms are x and 8, respectively. Hence, it makes sense to try the following.

\[x^{2}+16 x+64=(x+8)^{2} \nonumber \]

It is important that you check your result using multiplication. So, following the three-step algorithm for squaring a binomial:

- Square x to get \(x^2\).
- Multiply x and 8 to get 8x, then multiply this result by 2 to get 16x.
- Square 8 to get 64.

Hence, \(x^2 + 16x + 64\) is a perfect square trinomial and factors as \((x + 8)^2\).

As another example, consider \(x^2 − 10x + 25\). The square roots of the first and last terms are x and 5, respectively. Hence, it makes sense to try

\[x^{2}-10 x+25=(x-5)^{2} \nonumber \]

Again, you should check this result. Note especially that twice the product of x and −5 equals the middle term on the left, namely, −10x.

## Completing the Square

If a quadratic function is given in vertex form, it is a simple matter to sketch the parabola represented by the equation. For example, consider the quadratic function

\[f(x)=(x+2)^{2}+3 \nonumber \]

which is in vertex form. The graph of this equation is a parabola that opens upward. It is translated 2 units to the left and 3 units upward. This is the advantage of vertex form. The transformations required to draw the graph of the function are easy to spot when the equation is written in vertex form.

It’s a simple matter to transform the equation \(f(x) = (x + 2)^2 + 3\) into the general form of a quadratic function, \(f(x) = ax^2 + bx + c\). We simply use the three-step algorithm to square the binomial; then we combine like terms.

\[\begin{array}{l}{f(x)=(x+2)^{2}+3} \\ {f(x)=x^{2}+4 x+4+3} \\ {f(x)=x^{2}+4 x+7}\end{array} \nonumber \]

Note, however, that the result of this manipulation, \(f(x) = x^2 + 4x+ 7\), is not as useful as vertex form, as it is difficult to identify the transformations required to draw the parabola represented by the equation \(f(x) = x^2 + 4x + 7\).

It’s really the reverse of the manipulation above that is needed. If we are presented with an equation in the form \(f(x) = ax^2 + bx + c\), such as \(f(x) = x^2 + 4x + 7\), then an algebraic method is needed to convert this equation to vertex form \(f(x) = a(x−h)^2+k\); or in this case, back to its original vertex form \(f(x) = (x + 2)^2 + 3\).

The procedure we seek is called completing the square. The name is derived from the fact that we need to “complete” the trinomial on the right side of \(y = x^2 + 4x + 7\) so that it becomes a perfect square trinomial.

## Algorithm for Completing the Square

The procedure for completing the square involves three key steps.

- Take half of the coefficient of x and square the result.
- Add and subtract the quantity from step one so that the right-hand side of the equation does not change.
- Factor the resulting perfect square trinomial and combine constant terms.

Let’s follow this procedure and place \(f(x) = x^2 + 4x + 7\) in vertex form.

- Take half of the coefficient of x. Thus, (1/2)(4) = 2. Square this result. Thus, \(2^2 = 4\).
- Add and subtract 4 on the right side of the equation \(f(x) = x^2 + 4x + 7\) \[f(x)=x^{2}+4 x+4-4+7 \nonumber \]
- Group the first three terms on the right-hand side. These form a perfect square trinomial.

\[f(x)=\left(x^{2}+4 x+4\right)-4+7 \nonumber \]

Now factor the perfect square trinomial and combine the constants at the end to get

That’s it, we’re done! We’ve returned the general quadratic \(f(x) = x^2 + 4x + 7\) back to vertex form \(f(x) = (x + 2)^2 + 3\).

Let’s try that once more.

## Example \(\PageIndex{2}\)

Place the quadratic function \(f(x) = x^2 − 8x − 9\) in vertex form.

We follow the three-step algorithm for completing the square.

- Take half of the coefficient of x and square: i.e., \[[(1 / 2)(-8)]^{2}=[-4]^{2}=16 \nonumber \]
- Add and subtract this amount to the right-hand side of the function. \[f(x)=x^{2}-8 x+16-16-9 \nonumber \]
- Group the first three terms on the right-hand side. These form a perfect square trinomial. \[f(x)=\left(x^{2}-8 x+16\right)-16-9 \nonumber \]

Factor the perfect square trinomial and combine the coefficients at the end.

\[f(x)=(x-4)^{2}-25 \nonumber \]

Now, let’s see how we can use the technique of completing the square to help in drawing the graphs of general quadratic functions.

## Working with \(f(x) = x^2 + bx + c\)

The examples in this section will have the form \(f(x) = x^2 + bx + c\). Note that the coefficient of \(x^2\) is 1. In the next section, we will work with a harder form, \(f(x) = ax^2 + bx + c\), where \(a \neq 1\).

## Example \(\PageIndex{3}\)

Complete the square to place \(f(x) = x^2 + 6x + 2\) in vertex form and sketch its graph.

First, take half of the coefficient of x and square; i.e., \([(1/2)(6)]^2 = 9\). On the right side of the equation, add and subtract this amount so as to not change the equation.

\[f(x)=x^{2}+6 x+9-9+2 \nonumber \]

Group the first three terms on the right-hand side.

\[f(x)=\left(x^{2}+6 x+9\right)-9+2 \nonumber \]

The first three terms on the right-hand side form a perfect square trinomial that is easily factored. Also, combine the constants at the end.

\[f(x)=(x+3)^{2}-7 \nonumber \]

This is a parabola that opens upward. We need to shift the parabola 3 units to the left and then 7 units downward, placing the vertex at (−3, −7) as shown in Figure \(\PageIndex{1}\)(a). The axis of symmetry is the vertical line x = −3. The table in Figure \(\PageIndex{1}\)(b) calculates two points to the right of the axis of symmetry, and mirror points on the left of the axis of symmetry make for an accurate plot of the parabola.

Let’s look at another example.

## Example \(\PageIndex{4}\)

Complete the square to place \(f(x) = x^2 − 8x + 21\) in vertex form and sketch its graph.

First, take half of the coefficient of x and square; i.e., \([(1/2)(−8)]^2 = 16\). On the right side of the equation, add and subtract this amount so as to not change the equation.

\[f(x)=x^{2}-8 x+16-16+21 \nonumber \]

Group the first three terms on the right-hand side of the equation.

\[f(x)=\left(x^{2}-8 x+16\right)-16+21 \nonumber \]

The first three terms form a perfect square trinomial that is easily factored. Also, combine constants at the end.

\[f(x)=(x-4)^{2}+5 \nonumber \]

This is a parabola that opens upward. We need to shift the parabola 4 units to the right and then 5 units upward, placing the vertex at (4, 5), as shown in Figure \(\PageIndex{2}\)(a). The table in Figure \(\PageIndex{2}\)(b) calculates two points to the right of the axis of symmetry, and mirror points on the left of the axis of symmetry make for an accurate plot of the parabola.

## Working with \(f(x) = ax^2 + bx + c\)

In the last two examples, the coefficient of \(x^2\) was 1. In this section, we will learn how to complete the square when the coefficient of \(x^2\) is some number other than 1.

## Example \(\PageIndex{5}\)

Complete the square to place \(f(x)=2 x^{2}+4 x-4\) in vertex form and sketch its graph.

In the last two examples, we gained some measure of success when the coefficient of \(x^2\) was a 1. We were just getting comfortable with that situation and we’d like to continue to be comfortable, so let’s start by factoring a 2 from each term on the right-hand side of the equation.

\[f(x)=2\left[x^{2}+2 x-2\right] \nonumber \]

If we ignore the factor of 2 out front, the coefficient of \(x^2\) in the trinomial expression inside the parentheses is a 1. Ah, familiar ground! We will proceed as we did before, but we will carry the factor of 2 outside the parentheses in each step. Start by taking half of the coefficient of x and squaring the result; i.e., \([(1/2)(2)]^2 = 1\).

Add and subtract this amount inside the parentheses so as to not change the equation.

\[f(x)=2\left[x^{2}+2 x+1-1-2\right] \nonumber \]

Group the first three terms inside the parentheses and combine constants.

\[f(x)=2\left[\left(x^{2}+2 x+1\right)-3\right] \nonumber \]

The grouped terms inside the parentheses form a perfect square trinomial that is easily factored.

\[f(x)=2\left[(x+1)^{2}-3\right] \nonumber \]

Finally, redistribute the 2.

\[f(x)=2(x+1)^{2}-6 \nonumber \]

This is a parabola that opens upward. In addition, it is stretched by a factor of 2, so it will be somewhat narrower than our previous examples. The parabola is also shifted 1 unit to the left, then 6 units downward, placing the vertex at (−1, −6), as shown in Figure \(\PageIndex{3}\)(a). The table in Figure \(\PageIndex{3}\)(b) calculates two points to the right of the axis of symmetry, and mirror points on the left of the axis of symmetry make for an accurate plot of the parabola.

Let’s look at an example where the coefficient of \(x^2\) is negative.

## Example \(\PageIndex{6}\)

Complete the square to place \(f(x) = −x^2 + 6x − 2\) in vertex form and sketch its graph.

In the last example, we factored out the coefficient of \(x^2\). This left us with a trinomial having leading coefficient 1, which enabled us to proceed much as we did before: halve the middle coefficient and square, add and subtract this amount, factor the resulting perfect square trinomial. Since we were successful with this technique in the last example, let’s begin again by factoring out the leading coefficient, in this case a −1.

\[f(x)=-1\left[x^{2}-6 x+2\right] \nonumber \]

If we ignore the factor of −1 out front, the coefficient of \(x^2\) in the trinomial expression inside the parentheses is a 1. Again, familiar ground! We will proceed as we did before, but we will carry the factor of −1 outside the parentheses in each step. Start by taking half of the coefficient of x and squaring the result; i.e., \([(1/2)(−6)]^2 = 9\).

\[f(x)=-1\left[x^{2}-6 x+9-9+2\right] \nonumber \]

\[f(x)=-1\left[\left(x^{2}-6 x+9\right)-7\right] \nonumber \]

\[f(x)=-1\left[(x-3)^{2}-7\right] \nonumber \]

Finally, redistribute the −1.

\[f(x)=-(x-3)^{2}+7 \nonumber \]

This is a parabola that opens downward. The parabola is also shifted 3 units to the right, then 7 units upward, placing the vertex at (3, 7), as shown in Figure \(\PageIndex{4}\)(a). The table in Figure \(\PageIndex{4}\)(b) calculates two points to the right of the axis of symmetry, and mirror points on the left of the axis of symmetry make for an accurate plot of the parabola.

Let’s try one more example.

## Example \(\PageIndex{7}\)

Complete the square to place \(f(x) = 3x^2 + 4x − 8\) in vertex form and sketch its graph.

Let’s begin again by factoring out the leading coefficient, in this case a 3.

\[f(x)=3\left[x^{2}+\frac{4}{3} x-\frac{8}{3}\right] \nonumber \]

Fractions add a degree of difficulty, but, if you follow the same routine as in the previous examples, you should be able to get the needed result. Take half of the coefficient of x and square the result; i.e., \([(1/2)(4/3)]^2 = [2/3]^2 = 4/9\).

\[f(x)=3\left[x^{2}+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}-\frac{8}{3}\right] \nonumber \]

Group the first three terms inside the parentheses. You’ll need a common denominator to combine constants.

\[f(x)=3\left[\left(x^{2}+\frac{4}{3} x+\frac{4}{9}\right)-\frac{4}{9}-\frac{24}{9}\right] \nonumber \]

\[f(x)=3\left[\left(x+\frac{2}{3}\right)^{2}-\frac{28}{9}\right] \nonumber \]

Finally, redistribute the 3.

\[f(x)=3\left(x+\frac{2}{3}\right)^{2}-\frac{28}{3} \nonumber \]

This is a parabola that opens upward. It is also stretched by a factor of 3, so it will be narrower than all of our previous examples. The parabola is also shifted 2/3 units to the left, then 28/3 units downward, placing the vertex at (−2/3, −28/3), as shown in Figure \(\PageIndex{5}\)(a). The table in Figure \(\PageIndex{5}\)(b) calculates two points to the right of the axis of symmetry, and mirror points on the left of the axis of symmetry make for an accurate plot of the parabola.

In Exercises 1 - 8 , expand the binomial.

## Exercise \(\PageIndex{1}\)

\((x+\frac{4}{5})^2\)

\(x^2+\frac{8}{5}x+\frac{16}{25}\)

## Exercise \(\PageIndex{2}\)

\((x−\frac{4}{5})^2\)

## Exercise \(\PageIndex{3}\)

\((x+3)^2\)

\(x^2+6x+9\)

## Exercise \(\PageIndex{4}\)

\((x+5)^2\)

## Exercise \(\PageIndex{5}\)

\((x−7)^2\)

\(x^2−14x+49\)

## Exercise \(\PageIndex{6}\)

\((x−\frac{2}{5})^2\)

## Exercise \(\PageIndex{7}\)

\((x−6)^2\)

\(x^2−12x+36\)

## Exercise \(\PageIndex{8}\)

\((x−\frac{5}{2})^2\)

In Exercises 9 - 16 , factor the perfect square trinomial.

## Exercise \(\PageIndex{9}\)

\(x^2−\frac{6}{5}x+\frac{9}{25}\)

\((x−\frac{3}{5})^2\)

## Exercise \(\PageIndex{10}\)

\(x^2+5x+\frac{25}{4}\)

## Exercise \(\PageIndex{11}\)

Exercise \(\pageindex{12}\).

\(x^2+3x+\frac{9}{4}\)

## Exercise \(\PageIndex{13}\)

\(x^2+12x+36\)

\((x+6)^2\)

## Exercise \(\PageIndex{14}\)

\(x^2−\frac{3}{2}x+\frac{9}{16}\)

## Exercise \(\PageIndex{15}\)

\(x^2+18x+81\)

\((x+9)^2\)

## Exercise \(\PageIndex{16}\)

\(x^2+10x+25\)

In Exercises 17 - 24 , transform the given quadratic function into vertex form \(f(x) = (x−h)^2+k\) by completing the square.

## Exercise \(\PageIndex{17}\)

\(f(x) = x^2−x+8\)

\((x−\frac{1}{2})^2+\frac{31}{4}\)

## Exercise \(\PageIndex{18}\)

\(f(x) = x^2+x−7\)

## Exercise \(\PageIndex{19}\)

\(f(x) = x^2−5x−4\)

\((x−\frac{5}{2})^2−\frac{41}{4}\)

## Exercise \(\PageIndex{20}\)

\(f(x) = x^2+7x−1\)

## Exercise \(\PageIndex{21}\)

\(f(x) = x^2+2x−6\)

\((x+1)^2−7\)

## Exercise \(\PageIndex{22}\)

\(f(x) = x^2+4x+8\)

## Exercise \(\PageIndex{23}\)

\(f(x) = x^2−9x+3\)

\((x−\frac{9}{2})−\frac{69}{4}\)

## Exercise \(\PageIndex{24}\)

\(f(x) = x^2−7x+8\)

In Exercises 25 - 32 , transform the given quadratic function into vertex form \(f(x) = a(x−h)^2+k\) by completing the square.

## Exercise \(\PageIndex{25}\)

\(f(x) = −2x^2−9x−3\)

\(−2(x+\frac{9}{4})^2+\frac{57}{8}\)

## Exercise \(\PageIndex{26}\)

\(f(x) = −4x^2−6x+1\)

## Exercise \(\PageIndex{27}\)

\(f(x) = 5x^2+5x+5\)

\(5(x+\frac{1}{2})^2+\frac{15}{4}\)

## Exercise \(\PageIndex{28}\)

\(f(x) = 3x^2−4x−6\)

## Exercise \(\PageIndex{29}\)

\(f(x) = 5x^2+7x−3\)

\(5(x+\frac{7}{10})^2−\frac{109}{20}\)

## Exercise \(\PageIndex{30}\)

\(f(x) = 5x^2+6x+4\)

## Exercise \(\PageIndex{31}\)

\(f(x) = −x^2−x+4\)

\(−1(x+\frac{1}{2})^2+\frac{17}{4}\)

## Exercise \(\PageIndex{32}\)

\(f(x) = −3x^2−6x+4\)

In Exercises 33 - 38 , find the vertex of the graph of the given quadratic function.

## Exercise \(\PageIndex{33}\)

\(f(x) = −2x^2+5x+3\)

\((\frac{5}{4}, \frac{49}{8})\)

## Exercise \(\PageIndex{34}\)

\(f(x) = x^2+5x+8\)

## Exercise \(\PageIndex{35}\)

\(f(x) = −4x^2−4x+1\)

\((−\frac{1}{2}, 2)\)

## Exercise \(\PageIndex{36}\)

\(f(x) = 5x^2+7x+8\)

## Exercise \(\PageIndex{37}\)

\(f(x) = 4x^2+2x+8\)

\((−\frac{1}{4}, \frac{31}{4})\)

## Exercise \(\PageIndex{38}\)

In Exercises 39 - 44 , find the axis of symmetry of the graph of the given quadratic function.

## Exercise \(\PageIndex{39}\)

\(f(x) = −5x^2−7x−8\)

\(x = −\frac{7}{10}\)

## Exercise \(\PageIndex{40}\)

\(f(x) = x^2+6x+3\)

## Exercise \(\PageIndex{41}\)

\(f(x) = −2x^2−5x−8\)

\(x = −\frac{5}{4}\)

## Exercise \(\PageIndex{42}\)

\(f(x) = −x^2−6x+2\)

## Exercise \(\PageIndex{43}\)

\(f(x) = −5x^2+x+6\)

\(x = \frac{1}{10}\)

## Exercise \(\PageIndex{44}\)

\(f(x) = x^2−9x−6\)

For each of the quadratic functions in Exercises 45 - 66 , perform each of the following tasks.

- Use the technique of completing the square to place the given quadratic function in vertex form.
- Set up a coordinate system on a sheet of graph paper. Label and scale each axis.
- Draw the axis of symmetry and label it with its equation. Plot the vertex and label it with its coordinates.
- Set up a table near your coordinate system that calculates the coordinates of two points on either side of the axis of symmetry. Plot these points and their mirror images across the axis of symmetry. Draw the parabola and label it with its equation
- Use the graph of the parabola to determine the domain and range of the quadratic function. Describe the domain and range using interval notation.

## Exercise \(\PageIndex{45}\)

\(f(x) = x^2−8x+12\)

\(f(x) = (x−4)^2−4\)

Domain = \(\mathbb{R}\), Range = [−4, \(\infty\))

## Exercise \(\PageIndex{46}\)

\(f(x) = x^2+4x−1\)

## Exercise \(\PageIndex{47}\)

\(f(x) = (x+3)^2−6\)

Domain = \(\mathbb{R}\), Range = [−6, \(\infty\))

## Exercise \(\PageIndex{48}\)

\(f(x)=x^2−4x+1\)

## Exercise \(\PageIndex{49}\)

\(f(x) = x^2−2x−6\)

\(f(x) = (x−1)^2−7\)

Domain = \(\mathbb{R}\), Range = [−7, \(\infty\))

## Exercise \(\PageIndex{50}\)

\(f(x) = x^2+10x+23\)

## Exercise \(\PageIndex{51}\)

\(f(x) = −x^2+6x−4\)

\(f(x) = −(x−3)^2+5\)

Domain = \(\mathbb{R}\), Range = (−\(\infty\), 5]

## Exercise \(\PageIndex{52}\)

\(f(x) = −x^2−6x−3\)

## Exercise \(\PageIndex{53}\)

\(f(x) = −x^2−10x−21\)

\(f(x) = −(x+5)^2+4\)

Domain = \(\mathbb{R}\), Range = (−\(\infty\), 4]

## Exercise \(\PageIndex{54}\)

\(f(x) = −x^2+12x−33\)

## Exercise \(\PageIndex{55}\)

\(f(x) = 2x^2−8x+3\)

\(f(x) = 2(x−2)^2−5\)

Domain = \(\mathbb{R}\), Range = [−5, \(\infty\))

## Exercise \(\PageIndex{56}\)

\(f(x) = 2x^2+8x+4\)

## Exercise \(\PageIndex{57}\)

\(f(x) = −2x^2−12x−13\)

\(f(x) = −2(x+3)^2+5\)

## Exercise \(\PageIndex{58}\)

\(f(x) = −2x^2+24x−70\)

## Exercise \(\PageIndex{59}\)

\(f(x) = \frac{1}{2}x^2−4x+5\)

\(f(x) = \frac{1}{2}(x−4)^2−3\)

Domain = \(\mathbb{R}\), Range = [−3, \(\infty\))

## Exercise \(\PageIndex{60}\)

\(f(x) = \frac{1}{2}x^2+4x+6\)

## Exercise \(\PageIndex{61}\)

\(f(x) = −\frac{1}{2}x^2−3x+\frac{1}{2}\)

\(f(x) = −\frac{1}{2}(x+3)^2+5\)

## Exercise \(\PageIndex{62}\)

\(f(x) = −\frac{1}{2}x^2+4x−2\)

## Exercise \(\PageIndex{63}\)

\(f(x) = 2x^2+7x−2\)

\(f(x) = 2(x+\frac{7}{4})^2− \frac{65}{8}\)

Domain = \(\mathbb{R}\), Range = [\(−\frac{65}{8}\), \(\infty\))

## Exercise \(\PageIndex{64}\)

\(f(x) = −2x^2−5x−4\)

## Exercise \(\PageIndex{65}\)

\(f(x) = −3x^2+8x−3\)

\(f(x) = −3(x−\frac{4}{3})^2+\frac{7}{3}\)

Domain = \(\mathbb{R}\), Range = (−\(\infty\), \(\frac{7}{3}\)]

## Exercise \(\PageIndex{66}\)

\(f(x) = 3x^2+4x−6\)

In Exercises 67 - 72 , find the range of the given quadratic function. Express your answer in both interval and set notation.

## Exercise \(\PageIndex{67}\)

\(f(x) = −2x^2+4x+3\)

(\(−\infty\), 5] = {x|\(x \le 5\)}

## Exercise \(\PageIndex{68}\)

Exercise \(\pageindex{69}\).

\(f(x) = 5x^2+4x+4\)

[\(\frac{16}{5}\), \(\infty\)) = {x| \(x \ge 5\)}

## Exercise \(\PageIndex{70}\)

\(f(x) = 3x^2−8x+3\)

## Exercise \(\PageIndex{71}\)

\(f(x) = −x^2−2x−7\)

(\(−\infty\),−6] = {x|\(x \le −6\)}

## Exercise \(\PageIndex{72}\)

\(f(x) = x^2+x+9\)

Drill for Skill. In Exercises 73 - 76 , evaluate the function at the given value b.

## Exercise \(\PageIndex{73}\)

\(f(x) = 9x^2−9x+4\); b = −6

## Exercise \(\PageIndex{74}\)

\(f(x) = −12x^2+5x+2\); b = −3

## Exercise \(\PageIndex{75}\)

\(f(x) = 4x^2−6x−4\); b = 11

## Exercise \(\PageIndex{76}\)

\(f(x) = −2x^2−11x−10\); b = −12

Drill for Skill. In Exercises 77 - 80 , evaluate the function at the given expression.

## Exercise \(\PageIndex{77}\)

Evaluate f(x+4) if \(f(x) = −5x^2+4x+2\).

\(−5x^2−36x−62\)

## Exercise \(\PageIndex{78}\)

Evaluate f(−4x−5) if \(f(x) = 4x^2+x+1\).

## Exercise \(\PageIndex{79}\)

Evaluate f(4x−1) if \(f(x) = 4x^2+3x−3\).

\(64x^2−20x−2\)

## Exercise \(\PageIndex{80}\)

Evaluate f(−5x−3) if \(f(x) = −4x^2+x+4\).

## Choose Your Test

Sat / act prep online guides and tips, vertex form: what is it how do you calculate it.

General Education

Once you have the quadratic formula and the basics of quadratic equations down cold, it's time for the next level of your relationship with parabolas: learning about their vertex form .

Read on to learn more about the parabola vertex form and how to convert a quadratic equation from standard form to vertex form.

feature image credit: SBA73 /Flickr

## Why Is Vertex Form Useful? An Overview

The vertex form of an equation is an alternate way of writing out the equation of a parabola.

Normally, you'll see a quadratic equation written as $ax^2+bx+c$, which, when graphed, will be a parabola. From this form, it's easy enough to find the roots of the equation (where the parabola hits the $x$-axis) by setting the equation equal to zero (or using the quadratic formula).

If you need to find the vertex of a parabola, however, the standard quadratic form is much less helpful. Instead, you'll want to convert your quadratic equation into vertex form.

## What Is Vertex Form?

While the standard quadratic form is $ax^2+bx+c=y$, the vertex form of a quadratic equation is $\bi y=\bi a(\bi x-\bi h)^2+ \bi k$.

In both forms, $y$ is the $y$-coordinate, $x$ is the $x$-coordinate, and $a$ is the constant that tells you whether the parabola is facing up ($+a$) or down ($-a$). (I think about it as if the parabola was a bowl of applesauce; if there's a $+a$, I can add applesauce to the bowl; if there's a $-a$, I can shake the applesauce out of the bowl.)

The difference between a parabola's standard form and vertex form is that the vertex form of the equation also gives you the parabola's vertex: $(h,k)$.

For example, take a look at this fine parabola, $y=3(x+4/3)^2-2$:

Based on the graph, the parabola's vertex looks to be something like (-1.5,-2), but it's hard to tell exactly where the vertex is from just the graph alone. Fortunately, based on the equation $y=3(x+4/3)^2-2$, we know the vertex of this parabola is $(-4/3,-2)$.

Why is the vertex $(-4/3,-2)$ and not $(4/3,-2)$ (other than the graph, which makes it clear both the $x$- and $y$-coordinates of the vertex are negative)?

Remember: in the vertex form equation, $h$ is subtracted and $k$ is added . If you have a negative $h$ or a negative $k$, you'll need to make sure that you subtract the negative $h$ and add the negative $k$.

In this case, this means:

$y=3(x+4/3)^2-2=3(x-(-4/3))^2+(-2)$

and so the vertex is $(-4/3,-2)$.

You should always double-check your positive and negative signs when writing out a parabola in vertex form , particularly if the vertex does not have positive $x$ and $y$ values (or for you quadrant-heads out there, if it's not in quadrant I ). This is similar to the check you'd do if you were solving the quadratic formula ($x={-b±√{b^2-4ac}}/{2a}$) and needed to make sure you kept your positive and negatives straight for your $a$s, $b$s, and $c$s.

Below is a table with further examples of a few other parabola vertex form equations, along with their vertices. Note in particular the difference in the $(x-h)^2$ part of the parabola vertex form equation when the $x$ coordinate of the vertex is negative.

## How to Convert From Standard Quadratic Form to Vertex Form

Most of the time when you're asked to convert quadratic equations between different forms, you'll be going from standard form ($ax^2+bx+c$) to vertex form ($a(x-h)^2+k$).

The process of converting your equation from standard quadratic to vertex form involves doing a set of steps called completing the square. ( For more about completing the square, be sure to read this article .)

Let's walk through an example of converting an equation from standard form to vertex form. We'll start with the equation $y=7x^2+42x-3/14$.

The first thing you'll want to do is move the constant, or the term without an $x$ or $x^2$ next to it. In this case, our constant is $-3/14$. (We know it's negative $3/14$ because the standard quadratic equation is $ax^2+bx+c$, not $ax^2+bx-c$.)

First, we'll take that $-3/14$ and move it over to the left side of the equation:

$y+3/14=7x^2+42x$

The next step is to factor out the 7 (the $a$ value in the equation) from the right side, like so:

$y+3/14=7(x^2+6x)$

Great! This equation is looking much more like vertex form, $y=a(x-h)^2+k$.

At this point, you might be thinking, "All I need to do now is to move the $3/14$ back over to the right side of the equation, right?" Alas, not so fast.

If you take a look at part of the equation inside of the parentheses, you'll notice a problem: it's not in the form of $(x-h)^2$. There are too many $x$s! So we're not quite done yet.

What we need to do now is the hardest part—completing the square.

Let's take a closer look at the $x^2+6x$ part of the equation. In order to factor $(x^2+6x)$ into something resembling $(x-h)^2$, we're going to need to add a constant to the inside of the parentheses—and we're going to need to remember to add that constant to the other side of the equation as well (since the equation needs to stay balanced).

To set this up (and make sure we don't forget to add the constant to the other side of the equation), we're going to create a blank space where the constant will go on either side of the equation:

$y+3/14+7($ $)=7(x^2+6x+$ $)$

Note that on the left side of equation, we made sure to include our $a$ value, 7, in front of the space where our constant will go; this is because we're not just adding the constant to the right side of the equation, but we're multiplying the constant by whatever is on the outside of the parentheses. (If your $a$ value is 1, you don't need to worry about this.)

The next step is to complete the square. In this case, the square you're completing is the equation inside of the parentheses—by adding a constant, you're turning it into an equation that can be written as a square.

To calculate that new constant, take the value next to $x$ (6, in this case), divide it by 2, and square it.

$(6/2)^2=(3)^2=9$. The constant is 9.

The reason we halve the 6 and square it is that we know that in an equation in the form $(x+p)(x+p)$ (which is what we're trying to get to), $px+px=6x$, so $p=6/2$; to get the constant $p^2$, we thus have to take $6/2$ (our $p$) and square it.

Now, replace the blank space on either side of our equation with the constant 9:

$y+3/14+7(9)=7(x^2+6x+9)$

$y+{3/14}+63=7(x^2+6x+9)$

$y+{3/14}+{882/14}=7(x^2+6x+9)$

$y+{885/14}=7(x^2+6x+9)$

Next, factor the equation inside of the parentheses. Because we completed the square, you will be able to factor it as $(x+{\some \number})^2$.

$y+{885/14}=7(x+3)^2$

Last step: move the non-$y$ value from the left side of the equation back over to the right side:

$y=7(x+3)^2-{885/14}$

Congratulations! You've successfully converted your equation from standard quadratic to vertex form.

Now, most problems won't just ask you to convert your equations from standard form to vertex form; they'll want you to actually give the coordinates of the vertex of the parabola.

To avoid getting tricked by sign changes, let's write out the general vertex form equation directly above the vertex form equation we just calculated:

$y=a(x-h)^2+k$

And then we can easily find $h$ and $k$:

$+k=-{885/14}$

The vertex of this parabola is at coordinates $(-3,-{885/14})$.

Whew, that was a lot of shuffling numbers around! Fortunately, converting equations in the other direction (from vertex to standard form) is a lot simpler.

## How to Convert From Vertex Form to Standard Form

Converting equations from their vertex form to the regular quadratic form is a much more straightforward process: all you need to do is multiply out the vertex form.

Let's take our example equation from earlier, $y=3(x+4/3)^2-2$. To turn this into standard form, we just expand out the right side of the equation:

$$y=3(x+4/3)^2-2$$

$$y=3(x+4/3)(x+4/3)-2$$

$$y=3(x^2+{8/3}x+16/9)-2$$

$$y=3x^2+8x+{16/3}-2$$

$$y=3x^2+8x+{16/3}-{6/3}$$

$$y=3x^2+8x+10/3$$

Tada! You've successfully converted $y=3(x+4/3)^2-2$ to its $ax^2+bx+c$ form.

## Parabola Vertex Form Practice: Sample Questions

To wrap up this exploration of vertex form, we have four example problems and explanations. See if you can solve the problems yourself before reading through the explanations!

#1: What is the vertex form of the quadratic equation $x^2+ 2.6x+1.2$?

#2: Convert the equation $7y=91x^2-112$ into vertex form. What is the vertex?

#3: Given the equation $y=2(x-3/2)^2-9$, what are the $x$-coordinates of where this equation intersects with the $x$-axis?

#4: Find the vertex of the parabola $y=({1/9}x-6)(x+4)$.

## Parabola Vertex Form Practice: Solutions

#1: What is the vertex form of the quadratic equation ${\bi x^2}+ 2.6\bi x+1.2$?

Start by separating out the non-$x$ variable onto the other side of the equation:

$y-1.2=x^2+2.6x$

Since our $a$ (as in $ax^2+bx+c$) in the original equation is equal to 1, we don't need to factor it out of the right side here (although if you want, you can write $y-1.2=1(x^2+2.6x)$).

Next, divide the $x$ coefficient (2.6) by 2 and square it, then add the resulting number to both sides of the equation:

$(2.6/2)^2=(1.3)^2=1.69$

$y-1.2+1(1.69)=1(x^2+2.6x+1.69)$

Factor the right side of the equation inside the parentheses:

$y-1.2+1.69=(x+1.3)^2$

Finally, combine the constants on the left side of the equation, then move them over to the right side.

$y+0.49=(x+1.3)^2$

Our answer is $y=(x+1.3)^2-0.49$.

#2: Convert the equation $7\bi y=91\bi x^2-112$ into vertex form. What is the vertex?

When converting an equation into vertex form, you want the $y$ have a coefficient of 1, so the first thing we're going to do is divide both sides of this equation by 7:

$7y= 91x^2-112$

${7y}/7= {91x^2}/7-112/7$

$y=13x^2-16$

Next, bring the constant over to the left side of the equation:

$y+16=13x^2$

Factor out the coefficient of the $x^2$ number (the $a$) from the right side of the equation

$y+16=13(x^2)$

Now, normally you'd have to complete the square on the right side of the equation inside of the parentheses. However, $x^2$ is already a square, so you don't need to do anything besides moving the constant from the left side of the equation back to the right side:

$y=13(x^2)-16$.

Now to find the vertex:

$y=13(x^2)-16$

$-h=0$, so $h=0$

$+k=-16$, so $k=-16$

The vertex of the parabola is at $(0, -16)$.

#3: Given the equation $\bi y=2(\bi x-3/2)^2-9$, what is(are) the $\bi x$-coordinate(s) of where this equation intersects with the $\bi x$-axis?

Because the question is asking you to find the $x$-intercept(s) of the equation, the first step is to set $y=0$.

$y=0=2(x-3/2)^2-9$.

Now, there are a couple of ways to go from here. The sneaky way is to use the fact that there's already a square written into the vertex form equation to our advantage.

First, we'll move the constant over to the left side of the equation:

$0=2(x-3/2)^2-9$

$9=2(x-3/2)^2$

Next, we'll divide both sides of the equation by 2:

$9/2=(x-3/2)^2$

Now, the sneaky part. Take the square root of both sides of the equation:

$√(9/2)=√{(x-3/2)^2}$

$±3/{√2}=(x-3/2)$

Laura graduated magna cum laude from Wellesley College with a BA in Music and Psychology, and earned a Master's degree in Composition from the Longy School of Music of Bard College. She scored 99 percentile scores on the SAT and GRE and loves advising students on how to excel in high school.

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## VERTEX FORM OF A QUADRATIC FUNCTION WORKSHEET

1. Write the quadratic function in vertex form whose graph is shown below.

Questions 2-4 : Write the quadratic function in vertex form and write its vertex.

2. f(x) = x 2 + 6x + 9

3. f(x) = -x 2 - 2x - 2

4. f(x) = 4x 2 + 16x + 5

5. Write the following quadratic function in vertex form and sketch its graph :

f(x) = -x 2 + 2 x + 3

1. Answer :

The parabola above opens up with vertex (2, 1).

Vertex form of a quadratic function :

f(x) = a(x - h) 2 + k

Substitute vertex (h, k) = (2, 1).

f(x) = a(x - 2) 2 + 1 ----(1)

The y-intercept of the parabola is (0, 5). That is, the parabola is passing through (0, 5).

Substitute x = 0 and y = 5.

5 = a(0 - 2) 2 + 1

5 = a(- 2) 2 + 1

5 = 4a + 1

Subtract 1 from both sides.

Divide both sides by 4.

Substitute a = 1 in (1).

f(x) = 1(x - 2) 2 + 1

2. Answer :

Vertex form of the quadratic function :

f(x) = x 2 + 6x + 9

f(x) = x 2 + 2(x)(3) + 3 2

f(x) = (x + 3) 2

Comparing y = a(x - h) 2 + k and f(x) = (x + 3) 2 ,

h = -3 and k = 0

So, the vertex is (h, k) = (-3, 0).

3. Answer :

f(x) = -x 2 - 2 x - 2

f(x) = -(x 2 + 2x) - 2

f(x) = -[x 2 + 2(x)(1) + 1 2 - 1 2 ] - 2

f(x) = -[(x + 1) 2 - 1 2 ] - 2

f(x) = -[(x + 1) 2 - 1] - 2

f(x) = -(x + 1) 2 + 1 - 2

f(x) = -(x + 1) 2 - 1

Comparing f(x) = a(x - h) 2 + k and f(x) = -(x + 1) 2 - 1,

h = -1 and k = -1

So, the vertex is (h, k) = (-1, -1).

4. Answer :

f(x) = 4x 2 + 16x + 5

f(x) = 4(x 2 + 4x) + 5

f(x) = 4[x 2 + 2(x)(2) + 2 2 - 2 2 ] + 5

f(x) = 4[(x + 2) 2 - 2 2 ] + 5

f(x) = 4[(x + 2) 2 - 4] + 5

f(x) = 4(x + 2) 2 - 16 + 5

f(x) = 4(x + 2) 2 - 11

Comparing f(x) = a(x - h) 2 + k and f(x) = 4(x + 2) 2 - 11,

h = -2 and k = -11

So, the vertex is (h, k) = (-2, -11).

5. Answer :

Vertex form of the quadratic equation :

f(x) = -(x 2 - 2x) + 3

f(x) = -[x 2 - 2(x)(1) + 1 2 - 1 2 ] + 3

f(x) = -[(x - 1) 2 - 1 2 ] + 3

f(x) = -[(x - 1) 2 - 1] + 3

f(x) = -(x - 1) 2 + 1 + 3

f(x) = -(x - 1) 2 + 4

Comparing f(x) = a(x - h) 2 + k and f(x) = -(x - 1) 2 + 4,

h = 1 and k = 4

So, the vertex is (h, k) = (1, 4).

To graph the above quadratic equation, we need to find x-intercept and y-intercept, if any.

x-intercept :

To find the x-intercept, substitute f(x) = 0.

0 = -(x - 1) 2 + 4

(x - 1) 2 = 4

Take square root on both sides.

x - 1 = ±√4

x - 1 = ±2

x - 1 = -2 or x - 1 = 2

x = -1 or x = 3

So, the x-intercepts are (-1, 0) and (3, 0) .

y-intercept :

To find the y-intercept, substitute x = 0.

f(0) = -(0 - 1) 2 + 4

= -(-1) 2 + 4

= -1 + 4

So, the y-intercept is (0, 3).

Comparing f(x) = a(x - h) 2 + k and f(x) = -(x - 1) 2 + 4,

So, the parabola opens down with vertex (1, 4), x-intercepts (-1, 0) and (3, 0) and y-intercept (0, 3).

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Vertex Form of Parabolas Date_____ Period____ Use the information provided to write the vertex form equation of each parabola. 1) y = x2 + 16 x + 71 y = (x + 8 ... Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com. Title: Vertex Form of Parabolas

Worksheet by Kuta Software LLC FUNctions HW - Graphing Quadratic Functions in Vertex Form ... HW - Graphing Quadratic Functions in Vertex Form Name_____ ©g [2h0D1u7i QKFumtNaU zShoufctTwLaWrwej RLILhCo.b f PAVlGlc LrgiagAhOtasb lrQeBsxejrAvzegdO.-1- Sketch the graph of each function. ...

O Worksheet by Kuta Software LLC Algebra 2 ID: 1 Name_____ Date_____ Period____ ©R G2T0 b1o4I LK1untfa A iS ocfYtSwBajr 3e g kLQLiCF.X 4 UA 4lfl 4 yryitgjhatbs l drAeGswe5r9v7e1de. k Vertex Form Practice Use the information provided to write the vertex form equation of each parabola. 1) y = x2 − 4x + 5

Identify the vertex, axis of symmetry, and direction of opening of each. 1) y = 2(x + 10)2 + 1 ... Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com. Title: Graphing and Properties of Parabolas Author: Mike Created Date:

If a quadratic function is given in vertex form, it is a simple matter to sketch the parabola represented by the equation. For example, consider the quadratic function \[f(x)=(x+2)^{2}+3 \nonumber \] which is in vertex form. The graph of this equation is a parabola that opens upward. It is translated 2 units to the left and 3 units upward.

Any quadratic equation can be expressed in the form y = a(x-h)2+k. This is. called the vertex form of a quadratic equation. The graph of a quadratic equation forms a. parabola. The width, direction, and vertex of the parabola can all be found from this. equation.

Identify the vertex of each. 1) y = x2 + 16 x + 64 2) y = 2x2 − 4x − 2 ... Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com. Title: Properties of Parabolas Author: Mike Created Date: 7/25/2012 4:11:55 PM ...

Vertex form Graph quadratics in vertex form Google Classroom You might need: Calculator Graph the function. g ( x) = 1 3 ( x − 6) 2 + 1 1 2 3 4 5 6 7 8 9 − 2 − 3 − 4 − 5 − 6 − 7 − 8 − 9 1 2 3 4 5 6 7 8 9 − 2 − 3 − 4 − 5 − 6 − 7 − 8 − 9 y x Show Calculator Stuck? Do 4 problems

y = a (x-h)^2 + k is the vertex form equation. Now expand the square and simplify. You should get y = a (x^2 -2hx + h^2) + k. Multiply by the coefficient of a and get y = ax^2 -2ahx +ah^2 + k. This is standard form of a quadratic equation, with the normal a, b and c in ax^2 + bx + c equaling a, -2ah and ah^2 + k, respectively. 1 comment.

Now you are ready to create your Quadratic Functions and Inequalities Worksheet by pressing the Create Button. If You Experience Display Problems with Your Math Worksheet. This algebra 2 worksheet will produce problems for writing equations of parabolas. You may select different properties of the parabolas to be given for the problems.

Identify the vertex of the quadratic function in STANDARD form. Remember to use a b x 2 13. y 2 x2 16 x 31 14. y x2 4x 1 15. y 3x2 6x 4 Given a quadratic equation in vertex form, find the vertex, axis of symmetry, whether the graph opens up or down, the maximum or minimum, and the y-intercept. Graph it! 16. y 2 x 2 2 4 Opens:

Finding the vertex of the quadratic by using the equation x=-b/2a, and then substituting that answer for y in the orginal equation. Then, substitute the vertex into the vertex form equation, y=a (x-h)^2+k. (a will stay the same, h is x, and k is y). Also, remember that your h, when plugged into the equation, must be the additive inverse of what ...

9. Rewrite the equation of a parabola in vertex form. yx x = 2 16 59 10. Rewrite the equation of a parabola in vertex form. yx x = 2 23 11. Rewrite the equation of a parabola in vertex form. yx x = 2 616 12. Rewrite the equation of a parabola in vertex form. yx x = 2 10 34 13. Find p, q, and r so that this equation is a

Regents Exam Questions F.IF.C.8: Vertex Form of a Quadratic www.jmap.org F.IF.C.8: Vertex Form of a Quadratic In the function f(x) = (x − 2)2 + 4, the minimum value occurs when x is −2 2 −4 4 If Lylah completes the square for f(x) = x2 − 12x + 7 in order to find the minimum, she must write f(x) in the general form f(x) = (x − a)2 + b.

Axis of Symmetry: Function in standard form: Y‐int: Write the equation of the graph in vertex form. Compare f(x) = ‐ (x + 3)2 + 4 to the graph. Describe differences and similarities. 10.2 Quadratics in Vertex Form. PRACTICE.

While the standard quadratic form is a x 2 + b x + c = y, the vertex form of a quadratic equation is y = a ( x − h) 2 + k. In both forms, y is the y -coordinate, x is the x -coordinate, and a is the constant that tells you whether the parabola is facing up ( + a) or down ( − a ). (I think about it as if the parabola was a bowl of applesauce ...

1) Write an equation of the line through the points (2,-3) and (-1,0). Solve: 2 x − 5 = 3 3) Solve: 7 x − 3 ( x − 2 ) = 2( 5 − x ) 4) Solve the system : x − 2 y = 16 − 2x − y = −2 5) Solve the system: y = 2x + 7 4x - y = - 3 6) Find the x and y intercepts of the line 3y − x = 4 7) Evaluate: 3 x 2 − + 4 x when x = −2

2) the roots of the parabola can be found via the quadratic formula. So applying the arithmetic average formula (a+b)/2 where a is -b+sqrt (bsquared-4ac)/2a and b is -b-sqrt (bsquared-4ac)/a gives -b/2a as solution for x coordinate of vertex. This formula also works if the parabola has only one root.

to Vertex Form. 1) Find the vertex 2) Substitute , , and Vertex: (1, 3) into Converting from Vertex Form to Standard Form: Use the FOIL Method to find the product of the squared polynomial. Simplify using order of operations and arrange in descending order of power. Example: Convert to Standard Form. 1) FOIL Method

VERTEX FORM OF A QUADRATIC FUNCTION WORKSHEET. 1. Write the quadratic function in vertex form whose graph is shown below. Questions 2-4 : Write the quadratic function in vertex form and write its vertex. 2. f (x) = x 2 + 6x + 9. 3. f (x) = -x 2 - 2x - 2. 4. f (x) = 4x 2 + 16x + 5. 5.

Quadratic word problems (vertex form) Google Classroom. You might need: Calculator. Shenelle has 100 meters of fencing to build a rectangular garden. The garden's area (in square meters) as a function of the garden's width x (in meters) is modeled by: A ( x) = − ( x − 25) 2 + 625.

Quiz & Worksheet Goals. Take this quiz to test your ability to use the vertex form of a quadratic equation, including: Identifying the vertex in a vertex form. Identifying the vertex in a standard ...

Graphing Quadratics in Vertex Form Algebra 1 Practice Worksheet. Created by. Lisa Davenport. This 8-question algebra 1 worksheet provides students with organized practice graphing quadratics in Vertex Form. Students will first identify the vertex and then complete a table of values in order to graph the parabola.