Topics
7. Quadratic Functions
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United States of AmericaTennessee
Algebra 1

7.04 Quadratic functions in vertex form

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Concept summary

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)}
The coordinates of the vertex.
\bm{a}
The scale factor, which tells us about the shape of the graph.

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 x^2+bx+c\displaystyle =\displaystyle 0
2\displaystyle x^2+bx\displaystyle =\displaystyle -cSubtract c from both sides
3\displaystyle x^2+2\left(\frac{b}{2}\right)x\displaystyle =\displaystyle -cRewrite the x term
4\displaystyle x^2+2\left(\frac{b}{2}\right)x+\left(\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Add \left(\dfrac{b}{2}\right)^2 to both sides
5\displaystyle \left(x+\frac{b}{2}\right)^2\displaystyle =\displaystyle -c+\left(\frac{b}{2}\right)^2Factor the trinomial

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

Once our equation is in vertex form after completing the square, if we know the coordinates of the vertex, we actually only need to know one other point on the graph, such as the y-intercept, to be able to draw the graph of quadratic function. As the parabola is symmetric across the line of symmetry, which the vertex lies on, we can use the properties of symmetry to find a third point on the graph.

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

  • Vertex at \left(2, -3\right)
  • Axis of symmetry x=2
  • y-intercept at \left(0,1\right) has a corresponding point at \left(4, 1\right)

Worked examples

Example 1

The table of values below represents a quadratic function.

x-4-3-2-1012
p(x)-503430-5
a

Write the function p(x) in vertex form.

Approach

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

Solution

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), and therefore the quadratic is of the form 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

Using the equation found in part (a), predict the value of p(x) when x=6.

Approach

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

Solution

The equation found in part (a) is

\displaystyle p(x)\displaystyle =\displaystyle -(x+1)^2+4
\displaystyle p(x)\displaystyle =\displaystyle -(6+1)^2+4Substitute x=6 into the equation
\displaystyle p(x)\displaystyle =\displaystyle -(7)^2+4Simplify the parantheses
\displaystyle p(x)\displaystyle =\displaystyle -49+4Square what's inside the parantheses
\displaystyle p(x)\displaystyle =\displaystyle -45Simplify

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

Example 2

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).

Solution

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.

b

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

Approach

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).

Solution

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.

Example 3

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

a

Find the x-value of the vertex using the vertex formula.

Approach

The vertex formula to find the x-value of the vertex is h = \dfrac{-b}{2a}, where h is the x-value of the vertex, and a and b come from the standard form of a quadratic equation: y = ax^2 + bx + c.

Solution

We can start by identifying a and b in the given equation. a = 1,\,b=-4

Then we can substitute these values into the vertex formula:

\displaystyle h\displaystyle =\displaystyle \dfrac{-b}{2a}
\displaystyle h\displaystyle =\displaystyle \dfrac{-(-4)}{2 \times 1}Substitute in a = 1 and b = -4
\displaystyle h\displaystyle =\displaystyle 2Simplify the fraction

Therefore the x-value of the vertex is equal to 2.

Reflection

This also tells us that the axis of symmetry of the parabola is the line x=2.

b

Rewrite the equation in vertex form by completing the square.

Approach

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

Solution

We'll start with our original equationy = x^{2} - 4 x + 6

\displaystyle y\displaystyle =\displaystyle x^{2} - 4 x + (\frac{-4}{2})^{2} + 6 - (\frac{-4}{2})^{2}Complete the square by adding and subtracting (\frac{b}{2})^2
\displaystyle y\displaystyle =\displaystyle x^{2} - 4 x + 4 + 6 - 4 Simplify
\displaystyle y\displaystyle =\displaystyle (x - 2)^{2} + 6 - 4Factor the RHS
\displaystyle y\displaystyle =\displaystyle (x - 2)^{2} + 2Simplify

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

Reflection

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

c

Sketch the graph of the parabola.

Approach

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 find the x- and y-intercepts.

Solution

The y-value of the vertex is 2 since the vertex form of the equation is y=(x-2)^2 + 2. This means that the vertex is located at \left(2, 2 \right). We can also find the x- and y-intercepts by letting y and x=0 respectively.

\displaystyle y\displaystyle =\displaystyle (x-2)^2 + 2
\displaystyle 0\displaystyle =\displaystyle (x-2)^2 + 2Let y=0
\displaystyle -2\displaystyle =\displaystyle (x-2)^2 Subtract 2 from both sides

The next step would be to square root both sides of the equation to solve for x. However, we can't take the square root of a negative number. Therefore, there are no x-intercepts.

\displaystyle y\displaystyle =\displaystyle (x-2)^2 + 2
\displaystyle y\displaystyle =\displaystyle (0-2)^2 + 2Let x=0
\displaystyle y\displaystyle =\displaystyle 6Simplify

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

We can use the vertex and y-intercept to sketch the parabola.

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Outcomes

A1.N.Q.A.1

Use units as a way to understand real-world problems.*

A1.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.*

A1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A1.A.REI.D.5

Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line).

A1.F.IF.C.7

Graph functions expressed algebraically and show key features of the graph by hand and using technology.*

A1.F.IF.C.8

Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. *

A1.F.IF.C.8.A

Rewrite quadratic functions to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a real-world context.

A1.F.IF.C.9

Compare properties of functions represented algebraically, graphically, numerically in tables, or by verbal descriptions.*

A1.F.BF.B.2

Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given graphs.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP2

Reason abstractly and quantitatively.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP4

Model with mathematics.

A1.MP6

Attend to precision.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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