Topics
7. Quadratic Functions
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Algebra 1

7.03 Quadratic functions in vertex form

Lesson
Practice

Vertex form

Exploration

Use the orange, blue, and red sliders to change the quadratic functions.

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  1. What happens to the orange graph as the slider moves from a positive to a negative number?

  2. What happens to the blue graph as the slider moves from a positive to a negative number?

  3. What happens to the red graph as the slider moves from a positive to a negative number?

One way to represent quadratic functions is using vertex form. This form allows us to identify the coordinates of the vertex of the parabola, as well as the direction of opening and scale factor that compresses or stretches the graph of the function.

\displaystyle f\left(x\right) = a\left(x - h\right)^{2} + k
\bm{\left(h, k\right)}
coordinates of the vertex
\bm{a}
scale factor

When given the graph of the function, we can write the equation in vertex form using transformations.

A parabola can be vertically translated by increasing or decreasing the y-values by a constant number. A translation of y=x^{2} up by k units gives the equation y=x^{2}+k.

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\, \\This graph shows y=x^{2} translated vertically up by 2 to get y=x^{2} + 2, and down by 2 to get y=x^{2} - 2.

Similarly, a parabola can be horizontally translated by increasing or decreasing the x-values by a constant number. However, the x-value together with the translation must be squared together. That is, to translate y=x^{2} to the left by h units we get y=\left(x+h\right)^{2}.

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\, \\This graph shows y=x^{2} translated horizontally left by 2 to get y=\left(x+2\right)^{2} and right by 2 to get y=\left(x-2\right)^{2}.

A parabola can be dilated by multiplying every y-value by a constant number greater than 1. So to expand the parabola y=x^{2} by a scale factor of a we get y=ax^{2}. We can compress a parabola by using a scale factor between 0 and 1.

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This graph shows y=x^2 vertically expanded by a scale factor of 2 to get y=2x^{2} and compressed by a scale factor of 2 to get y=\dfrac{x^{2}}{2}.

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\, \\Finally, we can reflect a parabola across the x-axis by multiplying by -1. So to reflect y=x^{2} across the x-axis we get y=-x^{2}. Notice that reflecting will change the parabola from opening up to opening down.

The x-value of the vertex, h, represents the horizontal translation; the y-value, k, represents the vertical translation; and the leading coefficient, a, represents the shape of the parabola and the direction it opens. The parent function of a quadratic is f\left(x\right)=x^{2}, so writing these translations in function notation becomes af\left(x-h\right)+k=a\left(x-h\right)^{2}+k.

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\, \\As an example, consider the graph of \\y=\left(x-2\right)^{2}-3

  • Translation of the parent function 3 units down and 2 units right with a vertex at \left(2, -3\right)

  • Axis of symmetry x=2

  • y-intercept at \left(0,1\right) can be calculated by substituting x=0 into the function

When writing a quadratic function from its graph, we can begin by identifying the vertex and substituting these values into the function for h and k. Then, we can use another point on the parabola, like the y-intercept, to help us find the value of a.

Examples

Example 1

Consider the following function:m\left(x\right)=-\dfrac{1}{2}\left(x-2\right)^{2}+8

a

Find the vertex.

Worked Solution
Create a strategy

The function is given in vertex form y=a\left(x - h\right)^{2} + k where the vertex is the point \left(h,k\right).

Apply the idea

In the equation m\left(x\right)=-\dfrac{1}{2}\left(x-2\right)^{2}+8 the x-coordinate of vertex is h=2 and the y-coordinate of vertex is k=8.

With h=2 and k=8, the ordered pair of the vertex is at \left(2, 8\right).

b

State the domain.

Worked Solution
Create a strategy

To find the domain of m\left(x\right), we want to find all possible x-values for which m\left(x\right) could be graphed.

Apply the idea

We know that for a parabola, there are no restrictions on which x-values can be graphed as each side of the parabola continues infinitely in either x direction.

Domain: \{x \vert- \infty \lt x \lt \infty\}

c

State the range.

Worked Solution
Create a strategy

To find the range, we want to find all possible values of m\left(x\right). The vertex of a parabola affects the range of the function, as it will be the maximum or minimum value of m\left(x\right).

Apply the idea

This parabola opens down, so the y-value of the vertex is the maximum value of the function. The parabola continues infinitely in the negative y direction.

Range: \{y \vert y \leq 8\}

d

Draw the graph of the function.

Worked Solution
Create a strategy

The scale factor is -\dfrac{1}{2} which is negative, so the graph will open down. We can draw the graph through any three points that we know are on it.

Apply the idea
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Start by plotting the vertex, found in part (a), on the graph.

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We can solve for the x-intercepts by substituting m\left(x\right)=0 and solving for x.

\begin{aligned} 0 &=-\dfrac{1}{2}\left(x-2\right)^{2}+8 \\ -8 &=-\dfrac{1}{2}\left(x-2\right)^{2} \\ 16 &=\left(x-2\right)^{2} \\ \pm 4 &=x-2 \end{aligned}

\begin{aligned} & x=2+4 \text{ and } x=2-4 \\ &x=6 \text{ and } x=-2 \end{aligned}

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\, \\We can solve for the y-intercept by substituting x=0 and solving for m\left(x\right).

\begin{aligned} m\left(x\right) &=-\dfrac{1}{2}\left(0 - 2\right)^{2} + 8 \\ m\left(x\right) &=-\dfrac{1}{2}\left(-2\right)^{2} + 8 \\ m\left(x\right) &=-\dfrac{1}{2}\left(4\right) + 8 \\ m\left(x\right) &=-2 + 8 \\ m\left(x\right) &=6 \end{aligned}

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\, \\Then, we can get an additional point by reflecting the y-intercept across the axis of symmetry.

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\, \\To finish the drawing of the graph of the function, connect the points plotted to draw the parabola.

Reflect and check

Technically we only need 3 points to graph a parabola, but finding more points can make our graph more precise.

Example 2

The table of values below represents a quadratic function.

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p\left(x\right)-503430-5
a

Write the function p\left(x\right) in vertex form.

Worked Solution
Create a strategy

We can use the fact that a quadratic function has symmetry about its vertex to identify the location of the vertex from the table.

Apply the idea

Looking at the values of p\left(x\right), we can see that it has a maximum value of 4 and falls off symmetrically on either side. So we know that the vertex is the point \left(-1, 4\right), so we can use that to set up the equation: p\left(x\right) = a\left(x + 1\right)^{2} + 4

We can now find the value of a by substituting any other pair of values from the table, such as \left(0, 3\right). Doing so, we get 3 = a\left(0 + 1\right)^{2} + 4which we can solve to get a=-1.

So the quadratic function shown in the table of values is p\left(x\right) = -\left(x + 1\right)^{2} + 4

b

Determine the value of p\left(x\right) when x=6.

Worked Solution
Create a strategy

We can substitute x = 6 into the equation that was created in part (a) in order to predict what p\left(x\right) will be equal to.

Apply the idea

The equation found in part (a) is

\displaystyle p\left(x\right)\displaystyle =\displaystyle -\left(x+1\right)^{2}+4Original equation
\displaystyle p\left(6\right)\displaystyle =\displaystyle -\left(6+1\right)^{2}+4Substitute x=6
\displaystyle =\displaystyle -\left(7\right)^{2}+4Evaluate the addition
\displaystyle =\displaystyle -49+4Evaluate the exponent
\displaystyle =\displaystyle -45Evaluate the addition

This means that when x=6, p\left(x\right) is equal to -45.

c

Determine the x- and y-intercepts of the function.

Worked Solution
Create a strategy

Use the table of values to identify the x-intercepts when y=0 and the y-intercept when x=0.

Apply the idea

The x-intercepts occur at \left(-3,0\right) and \left(1,0\right).

The y-intercept occurs at \left(0,3\right).

Example 3

The quadratic function f\left(x\right) = 2x^{2} has been transformed to produce a new quadratic function g\left(x\right), as shown in the graph:

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a

Describe the transformation from f\left(x\right) to g\left(x\right).

Worked Solution
Apply the idea

The function g\left(x\right) has the same shape and size as f\left(x\right), but has been shifted to the left. Comparing the vertices of the two parabolas, we can see that this is a translation of 6 units to the left.

Reflect and check

We could confirm that g\left(x\right) is a horizontal shift left by evaluating f\left(x+6\right)=2\left(x+6\right)^{2}:

g\left(-8\right)=8 and f\left(-8+6\right)=2\left(-8+6\right)^{2}=2\left(-2\right)^{2}=2\left(4\right)=8

g\left(-6\right)=0 and f\left(-6+6\right)=2\left(-6+6\right)^{2}=2\left(0\right)^{2}=2\left(0\right)=0

g\left(-4\right)=8 and f\left(-4+6\right)=2\left(-4+6\right)^{2}=2\left(2\right)^{2}=2\left(4\right)=8

b

Write the equation of the function g\left(x\right) in vertex form.

Worked Solution
Create a strategy

Remember that vertex form for a quadratic is g\left(x\right) = a\left(x - h\right)^2 + k, where the vertex is at the point \left(h, k\right).

Apply the idea

f\left(x\right) = 2x^{2} has been translated 6 units to the left to produce g\left(x\right), and we can see that its vertex is at \left(-6, 0\right). So g\left(x\right) has can be written as g\left(x\right) = 2\left(x + 6\right)^{2}.

c

Describe the transformations of the graph of f\left(x\right) resulting in the function h\left(x\right)=-2\left(x+6\right)^{2} +3.

Worked Solution
Create a strategy

Since f\left(x\right) opens upward and has a vertex at \left(0,0\right), the vertex form of h\left(x\right) gives information about its vertex and direction of its opening.

Apply the idea

The function h\left(x\right) will open downward and be a reflection of f\left(x\right) across the x-axis, since the value of a became negative.

The vertex of h\left(x\right) is \left(-6, 3\right). The vertex became a maximum value and shifted the graph up 3 units and to the left 6 units.

d

Sketch the graph of h\left(x\right).

Worked Solution
Create a strategy

Use the description of the transformations of g\left(x\right) from part (c) and graph of g\left(x\right) to sketch h\left(x\right).

Apply the idea

Start by reflecting g\left(x\right) across the x-axis using points from its graph.

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Perform a vertical shift 3 units up.

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The graph of h\left(x\right) follows:

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Example 4

Meri throws a rock into Crescent Lake. The height of the rock above ground is a quadratic function of time. The rock is thrown from 4.4 \text{ ft} above ground. After 1.5 seconds, the rock reaches its maximum height of 24 \text{ ft}. Write the quadratic equation in vertex form.

Worked Solution
Create a strategy

The maximum height of the rock at 24 feet indicates that this is where the vertex of the function is located. This occurs at 1.5 seconds. Substitute the vertex of the graph and the y-intercept into the vertex form of a quadratic function to detrmine the equation.

Apply the idea

The vertex is located at \left(1.5, 24\right), so we can substitute h=1.5 and k=24.

The rock is thrown from 4.4 feet, so the y-intercept is located at \left(0,4.4\right) and we can substitute x=0 and y=4.4.

\displaystyle y\displaystyle =\displaystyle a\left(x-h\right)^{2}+kVertex form of a quadratic function
\displaystyle 4.4\displaystyle =\displaystyle a\left(0-1.5\right)^{2}+24Substitute x=0, y=4.4, h=1.5, and k=24
\displaystyle -19.6\displaystyle =\displaystyle a\left(-1.5\right)^{2}Subtract 24 from both sides
\displaystyle -19.6\displaystyle =\displaystyle a(2.25)Evaluate the exponent
\displaystyle -8.7\displaystyle =\displaystyle aDivide by 2.25 on both sides

The quadratic equation in vertex form that models the rock's height above the ground as a function of time is y=-8.7\left(x-1.5\right)^{2}+24.

Idea summary

The vertex form of a quadratic function is:

\displaystyle f\left(x\right) = a\left(x - h\right)^{2} + k
\bm{\left(h, k\right)}
coordinates of the vertex
\bm{a}
scale factor

Changing the values of a, h, and k will transform the graph in different ways:

  • a: vertical stretch or compression

  • h: horizontal translation

  • k: vertical translation

Completing the square

Exploration

  • Drag the blue slider to the right until the number 23 is shown in blue

  • Drag the red slider to the right to create the largest square possible without any missing pieces.

  • Check the box "Complete the square"

  • Repeat this process by changing the value of the blue and red sliders to any value of your choice

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  1. What patterns do you notice when working through the process of completing the square?

Completing the square is a method we use to rewrite a standard quadratic expression in vertex form.

Completing the square allows us to rewrite our equation so that it contains a perfect square trinomial. A perfect square trinomial takes on the form A^{2}+2AB+B^{2}=\left(A+B\right)^{2}, which is the same format we need to have an equation in vertex form.

For quadratic equations where a=1, we can write them in perfect square form by following these steps:

1\displaystyle y\displaystyle =\displaystyle x^{2}+bx+c
2\displaystyle y\displaystyle =\displaystyle x^{2}+2\left(\frac{b}{2}\right)x+cRewrite the x term
3\displaystyle y\displaystyle =\displaystyle x^{2}+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2}\right)^{2}+c-\left(\frac{b}{2}\right)^{2}Add and subtract \left(\dfrac{b}{2}\right)^{2} to keep the equation balanced
4\displaystyle y\displaystyle =\displaystyle \left(x+\frac{b}{2}\right)^{2}+c-\left(\frac{b}{2}\right)^{2}Factor the perfect square trinomial
5\displaystyle y\displaystyle =\displaystyle \left(x-h\right)^{2}+kMatch the completed square to vertex form

If a \neq 1, we can first divide through by a to factor it out.

The quadratic equation, when rewritten by completing the square, becomes the vertex form of a quadratic equation.

Examples

Example 5

Consider the following equation:y = x^{2} - 4 x + 6

a

Rewrite the equation in vertex form by completing the square.

Worked Solution
Create a strategy

We'll follow the standard complete the square method and stop working once our equation is in vertex form, y = a\left(x - h\right)^{2} + k.

Apply the idea
\displaystyle y\displaystyle =\displaystyle x^{2} - 4 x + 6Original equation
\displaystyle y\displaystyle =\displaystyle x^{2} - 4 x + \left(\frac{-4}{2}\right)^{2} + 6 - \left(\frac{-4}{2}\right)^{2}Add and subtract \left(\frac{b}{2}\right)^{2}
\displaystyle y\displaystyle =\displaystyle x^{2} - 4 x + 4 + 6 - 4 Evaluate the exponents
\displaystyle y\displaystyle =\displaystyle \left(x - 2\right)^{2} + 6 - 4Factor x^{2} - 4 x + 4
\displaystyle y\displaystyle =\displaystyle \left(x - 2\right)^{2} + 2Evaluate the subtraction

We've completed the square and the equation is now in vertex form.

Reflect and check

We must add and subtract \left(\dfrac{b}{2}\right)^{2} to the RHS of the equation so that the value of the equation does not change.

b

Determine the vertex of the quadratic function and if it is a minimum or maximum.

Worked Solution
Create a strategy

Use the vertex form of the quadratic function from part (a) to determine its vertex and whether it is a minimum or maximum value.

Apply the idea

The quadratic function in vertex form is y=\left(x-2\right)^{2} +2, meaning the vertex is located at the point \left(2,2\right).

Since the value of a is 1, we know that the graph opens upward and the vertex is a minimum value.

c

Sketch the graph of the parabola.

Worked Solution
Create a strategy

We can find key points of our parabola to sketch it. In part (a) we found the x-value of the vertex, and in part (b) we found the vertex form of our equation which shows us the y-value of our vertex. We can also use the vertex form to consider the x- and y-intercepts.

Apply the idea

The y-value of the vertex is 2 since the vertex form of the equation is y=\left(x-2\right)^{2} + 2. This means that the vertex is located at \left(2,2\right). We know the scale factor a is 1, so the parabola opens upward. Since the vertex is above the x-axis and opens up, the graph will not cross the x-axis and there are no x-intercepts.

Find the y-intercept:

\displaystyle y\displaystyle =\displaystyle \left(x-2\right)^{2} + 2Vertex form of the equation
\displaystyle y\displaystyle =\displaystyle \left(0-2\right)^{2} + 2Substitute x=0
\displaystyle y\displaystyle =\displaystyle 6Evaluate

Therefore the y-intercept is \left(0, 6\right).

We can use the vertex and y-intercept to sketch the parabola, remembering there is an axis of symmetry at the vertex.

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Idea summary

A quadratic function in standard form can be converted to vertex form by completing the square:

1\displaystyle y\displaystyle =\displaystyle x^{2}+bx+c
2\displaystyle y\displaystyle =\displaystyle x^{2}+2\left(\frac{b}{2}\right)x+cRewrite the x term
3\displaystyle y\displaystyle =\displaystyle x^{2}+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2}\right)^{2}+c-\left(\frac{b}{2}\right)^{2}Add and subtract \left(\dfrac{b}{2}\right)^{2} to keep the equation balanced
4\displaystyle y\displaystyle =\displaystyle \left(x+\frac{b}{2}\right)^{2}+c-\left(\frac{b}{2}\right)^{2}Factor the perfect square trinomial
5\displaystyle y\displaystyle =\displaystyle \left(x-h\right)^{2}+kMatch the completed square to vertex form

If a \neq 1, we can first divide through by a to factor it out.

Outcomes

A.F.2

The student will investigate, analyze, and compare characteristics of functions, including quadratic and exponential functions, and model quadratic and exponential relationships.

A.F.2b

Given an equation or graph, determine key characteristics of a quadratic function including x-intercepts (zeros), y-intercept, vertex (maximum or minimum), and domain and range (including when restricted by context); interpret key characteristics as related to contextual situations, where applicable.

A.F.2c

Graph a quadratic function, f(x), in two variables using a variety of strategies, including transformations f(x) + k and kf(x), where k is limited to rational values.

A.F.2g

For any value, x, in the domain of f, determine f(x) of a quadratic or exponential function. Determine x given any value f(x) in the range of f of a quadratic function. Explain the meaning of x and f(x) in context.

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