Up till now, we have focused on the vertex form of a quadratic function, [latex]f(x)=a(x-h)^2+k[/latex]. This form allows us to immediately identify the vertex [latex](h, k)[/latex]. If we were to multiply out this function we would end up with the standard form of a quadratic function, [latex]f(x)=ax^2+bx+c[/latex]. When a quadratic equation is written in standard form, the vertex is not obvious.

For example, the function [latex]f(x)=2(x-3)^2+1[/latex] has a vertex at (3, 1). But if we multiply out the function we get:

[latex]\begin{aligned}f(x)&=2(x-3)^2+1\\&=2(x-3)(x-3)+1\\&=2(x^2-6x+9)+1\\&=2x^2-12x+19\end{aligned}[/latex]

The vertex of the equivalent function [latex]f(x)=2x^2-12x+19[/latex] is no longer obvious.

This section considers three methods of determining the vertex when a quadratic function is given in the standard form [latex]f(x)=ax^2+bx+c[/latex].

Completing the Square

A common way to find the vertex of a quadratic function [latex]f(x)=ax^2+bx+c[/latex] is to convert the function from standard form into the vertex form [latex]f(x)=a(x-h)^2+k[/latex], where the vertex is [latex](h, k)[/latex]. This method requires us to form a perfect square on the [latex]x[/latex]-terms, [latex]ax^2+bx[/latex]. The constant [latex]c[/latex] will become part of the constant term [latex]k[/latex] in the vertex form.

So our goal is to turn [latex]f(x)=ax^2+bx+c[/latex] into [latex]f(x)=a(x-h)^2+k[/latex] by creating a perfect square out of [latex]ax^2+bx[/latex].

The Leading coefficient [latex]a=1[/latex]

Let’s start by considering the case when the leading coefficient [latex]a=1[/latex]. When the perfect square [latex](x+A)^2[/latex] is multiplied out we get,

[latex]\begin{aligned}(x+A)^2&=(x+A)(x+A)\\&=x^2+Ax+Ax+A^2\\&=x^2+2Ax+A^2\end{aligned}[/latex]

The middle term is doubled. So to go backwards and make a perfect square out of [latex]x^2+2Ax+A^2[/latex] we take half of the coefficient of [latex]x[/latex] inside the perfect square, [latex](x+A)^2[/latex].

For example, if we want to make a perfect square out of [latex]x^2+6x[/latex] we put half of +6 in with [latex]x[/latex] to get [latex](x+3)^2[/latex]. However, [latex](x+3)^2=x^2+6x+9[/latex]; we have an extra 9 that was not in our original expression [latex]x^2+6x[/latex]. But we can easily remove that extra 9 by subtracting it:

[latex]x^2+6x=(x+3)^2-9[/latex]

So, to complete the square on [latex]x^2+bx[/latex], we take half of the coefficient of [latex]x[/latex] and put it in the perfect square. But, we also have to subtract the extra term:

[latex]x^2+bx=\left(x+\dfrac{b}{2}\right)^2\color{blue}{-\left(\dfrac{b}{2}\right)^2}[/latex]

This process of making a perfect square and subtracting the extra term is called completing the square.

Suppose this time we want to complete the square on [latex]f(x)=x^2-8x+3[/latex]. Half of –8 is –4, so we get [latex](x-4)^2[/latex] as our perfect square. But [latex](x-4)^2=x^2-8x+16[/latex] so we need to subtract the extra term 16 from the original constant 3:

[latex]\begin{aligned}f(x)&=\color{blue}{x^2-8x}+3\\&=\color{blue}{(x-4)^2-16}+3\\&=(x-4)^2-13\end{aligned}[/latex]

Once we have converted a function into vertex form [latex]f(x)=a(x-h)^2+k[/latex] by completing the square, we can pick out the vertex. In this example, the vertex is (4, –13).

Remember that when we have completed the square, we can always check our answer by multiplying out the vertex form to make sure we get back to the standard form. In this example,

[latex]\begin{aligned}f(x)&=(x-4)^2-13\\&=(x-4)(x-4)-13\\&=x^2-4x-4x+16-13\\&=x^2-8x+3\end{aligned}[/latex]

We get back to where we started, so we know the completed square is correct.

The Leading Coefficient [latex]a\neq1[/latex]

In math, we often build on what we have already done, or use what we have already proved to extend the concept. We have just learned how to complete the square on a quadratic function where the coefficient is 1. When the leading coefficient is not 1, we can make it seem like it is 1 by factoring out the leading coefficient from the two terms that contain [latex]x[/latex] so that [latex]f(x)=ax^2+bx+c = a\left(x^2+\dfrac{b}{a}x\right)+c[/latex]. We don’t have to factor the constant term [latex]c[/latex] because it is not part of the perfect square and will be combined with the subtracted extra term to create [latex]k[/latex]. Now we can use the same method we just earned to complete the square inside the parentheses of the function.

Like before, this is easier to see with a specific example.

If you prefer, we can factor the leading coefficient from all terms in the function, including the constant. The next example shows that the only difference is that we will combine the constants before distributing the leading coefficient.

Using the Line of Symmetry to Find the Vertex

In 4.1, we introduced the idea of using two symmetric points to find the [latex]x[/latex]-coordinate of the vertex. The [latex]x[/latex]-coordinate of the vertex is the midpoint between any two symmetric points on the parabola. This is because the two symmetric points on a parabola are the same distance away from the line of symmetry (or from the [latex]x[/latex]-coordinate of the vertex). Once the [latex]x[/latex]-coordinate of the vertex is identified, we can determine the [latex]y[/latex]-coordinate by evaluating the function value at this coordinate.

Two symmetrical points on a parabola can be found by graphing or by setting the function [latex]f(x)=ax^2+bx+c[/latex] equal to a constant. The easiest constant to use is [latex]c[/latex] because the equation will simplify to an easily solved quadratic equation:

[latex]\begin{aligned}ax^2+bx+c &= c\\ax^2+bx&=0\end{aligned}[/latex]

Graphically, this equation is equivalent to finding the intersection points between the parabola representing the function with the horizontal line [latex]y = c[/latex] (figure 1). The two intersection points are symmetrical because the horizontal line is perpendicular to the line of symmetry of the parabola, and the parabola is symmetrical. Once the equation [latex]ax^2+bx+c = c[/latex] is set up, we can solve for [latex]x[/latex], where the [latex]x[/latex] values are the [latex]x[/latex]-coordinates of the two symmetric points.

Figure 1. Intersection of [latex]f(x)=ax^2+bx+c[/latex] and [latex]y=c[/latex]

Notice from figure 1 that solving [latex]ax^2+bx+c = c[/latex] will always result in the [latex]y[/latex]-intercept [latex](0, c)[/latex] being one of the symmetric points.

Suppose we are given the function [latex]f(x) = x^2 - 10x + 33[/latex]. To find two symmetric points, we may set the function equal to 33 (the value of [latex]c[/latex]). This equation, [latex]x^2-10x+33=33[/latex], will determine the two intersection points between the function[latex]f(x) = x^2 - 10x + 33[/latex] and the line [latex]y = 33[/latex]. The two intersection points are symmetrical with respect to the line of symmetry of the parabola of the function.

[latex]\begin{aligned}x^2 - 10x + 33 &= 33\\x^2 - 10x &= 0\\x(x - 10) &= 0\\x &= 0,\;10\end{aligned}[/latex]

SInce these points lie on the line [latex]y=33[/latex], their [latex]y[/latex]-coordinates are 33. Therefore, the two symmetric points are (0, 33), the [latex]y[/latex]-intercept of the function, and (10, 33). We can check this by finding the corresponding function values at [latex]x=0[/latex] and [latex]x=10[/latex]:

[latex]f(0)=(0)^2-10(0)+33=33[/latex] and [latex]f(10)=(10)^2-10(10)+33=33[/latex]

To find the axis of symmetry, and hence the [latex]x[/latex]-coordinate of the vertex, we only need the [latex]x[/latex]-coordinates of the two symmetric points. The [latex]x[/latex]-coordinate of the vertex is exactly halfway between [latex]x=0[/latex] and [latex]x=10[/latex]:

[latex]x=\dfrac{0+10}{2} = 5[/latex]

The [latex]y[/latex]-coordinate of the vertex is [latex]f(5)[/latex]:

[latex]\begin{aligned}f(5) &= (5)^2 - 10(5) + 33 \\&= 25 - 50 + 33 \\&= 8\end{aligned}[/latex]

Therefore, the vertex of the function is (5, 8).

 

Using a Formula to Find the Vertex

Building on what we have just learned, we can determine the vertex of a parabola that represents the function [latex]f(x)=ax^2+bx+c[/latex].

First we find two symmetric points by finding the intersection points of [latex]y=f(x)[/latex] and [latex]y=c[/latex]:

[latex]\begin{aligned}ax^2+bx+c&=c\\ax^2+bx&=0\\ax\left(x+\dfrac{b}{a}\right)&=0\\x&=0,\;-\dfrac{b}{a}\end{aligned}[/latex]

The [latex]x[/latex]-coordinate of the vertex lies exactly halfway between [latex]x=0[/latex] and [latex]x=-\dfrac{b}{a}[/latex]:

[latex]\dfrac{0+\frac{-b}{a}}{2}=-\dfrac{b}{2a}[/latex]

To find the [latex]y[/latex]-coordinate, we can then evaluate [latex]f\left(\dfrac{b}{2a}\right)[/latex].

Consequently, the vertex is [latex]\left(-\dfrac{b}{2a},\;f\left(-\dfrac{b}{2a}\right)\right)[/latex].

For example, given the function [latex]f(x) = x^2 - 10x + 33[/latex], the [latex]x[/latex]-coordinate of the vertex is:

[latex]\begin{aligned}x&=-\dfrac{b}{2a}\\&=-\dfrac{-10}{2(1)}\\&=5\end{aligned}[/latex]

The [latex]y[/latex]-coordinate of the vertex is:

 [latex]\begin{aligned}f(5)&= (5)^2 - 10(5) + 33 \\&= 25 - 50 +33 \\&= 8 \end{aligned}[/latex]

Therefore, the vertex of the parabola is (5, 8).

We now have three ways in which to algebraically determine the vertex of a parabola:

1. We can complete the square to convert the function from standard form [latex]f(x)=ax^2+bx+c[/latex] to vertex form [latex]f(x)=a(x-h)^2+k[/latex], then pick out the vertex [latex](h, k)[/latex].
2. We can find two symmetric points by determining the intersection points of [latex]f(x)=ax^2+bx+c[/latex] with the horizontal line [latex]y=c[/latex], then find the midpoint between the [latex]x[/latex]=coordinates of the points of intersection. This gives us the [latex]x[/latex]-coordinate of the vertex, and we can finds the [latex]y[/latex]-coordinate by evaluating the function at the determined [latex]x[/latex]-coordinate.
3. We can determine the [latex]x[/latex]-coordinate of the vertex using the formula [latex]x=-\dfrac{b}{2a}[/latex], then evaluate the function value [latex]f\left(-\dfrac{b}{2a}\right)[/latex] to determine the [latex]y[/latex]-coordinate.