Vertex Form

Section8.4Vertex Form

The graphing form of the equation of a parabola is

\begin{equation*} y=a(x-h)^2+k \end{equation*}

where $a\text{,}$ $h\text{,}$ and $k$ are all real numbers, $a \ne 0\text{.}$

When written in graphing form, we can infer the vertex of the parabola directly from the equation, it is the point $(h,k)\text{.}$ This fact is actually somewhat intuitive. No matter what values you choose for $x\text{,}$ the smallest value you'll ever get for the expression $(x-h)^2$ is zero, and that will occur when $x=h\text{.}$ So if $a$ is positive, the expression $a(x-h)^2+k$ will have its least value when $x=h$ and if the value of $a$ is negative, the expression $a(x-h)^2+k$ will have its greatest value when $x=h\text{.}$ So for any non-zero value of $a\text{,}$ the parabola will have either its lowest point or its highest point when $x=h\text{.}$ That low or high point is precisely the vertex. The fact that the $y$-coordinate has a value of $k$ when $x=h$ comes directly from substitution.

\begin{align*} y\amp=a(\highlight{h}-h)^2+k\\ \amp=a(0)^2+k\\ \amp=k \end{align*}

For example, given the equation

\begin{equation*} y=\frac{1}{2}(x-3)^2-7 \end{equation*}

we know immediately that the vertex of the parabola is the point $(3,-7)\text{.}$ Sometimes folks get a little uncertain about the sign of the value $h\text{.}$ If you start second guessing yourself, try to remember that $h$ is the value that causes $(x-h)^2$ to equal zero. In the last equation that gives us the following.

\begin{align*} x-3\amp=0\\ x-3\addright{3}\amp=0\addright{3}\\ x\amp=3 \end{align*}

Direct substitution gives us the $y$-coordinate when $x=3\text{.}$

\begin{align*} y\amp=\frac{1}{2}(\highlight{3}-3)^2-7\\ \amp=\frac{1}{2}(0)^2-7\\ \amp=-7 \end{align*}

The other information about the parabola is determined in the same way that it was determined when we worked with the standard form of the equation. To summarize:

When presented with an equation of form $y=a(x-h)^2+k\text{,}$ $a \ne 0\text{,}$ the equation graphs to a parabola with the following properties.

The vertex of the parabola is the point $(h,k)\text{.}$
The axis of symmetry for the parabola is the vertical line with equation $x=h\text{.}$
The parabola is concave up (opens upward) when $a \gt 0$ and the parabola is concave down (opens downward) when $a \lt 0\text{.}$
If there are $x$-intercept(s) on the parabola, the $x$-coordinates of those (or that) point(s) are the real number solutions to the equation $a(x-h)^2+k=0\text{.}$ When that equation has no real number solutions, the parabola has no $x$-intercepts.
The $y$-coordinate of the $y$-intercept is determined by replacing $x$ with zero in the equation $y=a(x-h)^2+k\text{.}$

Example8.4.1

Determine the vertex, axis of symmetry, concavity, and all intercepts for the parabola with equation

\begin{equation*} y=-2(x+6)^2+8\text{.} \end{equation*}

Solution

he vertex is the point $(-6,8)$ and the axis of symmetry is the vertical line with equation $x=-6\text{.}$ Note that $(x+6)^2$ is equivalent to $(x-(\highlight{-6}))\text{,}$ which is another way of seeing that the $x$-coordinate of the vertex is $-6\text{.}$

Because the value of $a\text{,}$ $-2\text{,}$ is negative, we know that the parabola is concave down.

To determine the $x$-coordinates of the $x$-intercepts, we need to solve the equation $-2(x+6)^2+8=0\text{.}$ Because of the form of this equation, by far the fastest route to solution is the square root method.

\begin{align*} -2(x+6)^2+8\amp=0\\ -2(x+6)^2+8\subtractright{8}\amp=0\subtractright{8}\\ -2(x+6)^2\amp=-8\\ \divideunder{-2(x+6)^2}{-2}\amp=\divideunder{-8}{-2}\\ (x+6)^2\amp=4\\ x+6\amp=\pm\sqrt{4}\\ x+6\amp=\pm 2 \end{align*}

\begin{align*} x+6\amp=-2 \amp\amp\text{ or }\amp x+6\amp=2\\ x+6\subtractright{6}\amp=-2\subtractright{6} \amp\amp\text{ or }\amp x+6\subtractright{6}\amp=2\subtractright{6}\\ x\amp=-8 \amp\amp\text{ or }\amp x\amp=-4 \end{align*}

The $x$-intercepts on the parabola are $(-8,0)$ and $(-4,0)\text{.}$ Note that the parabola having its an axis of symmetry with equation $x=-6$ is consistent with these intercepts — their $x$-coordinates are equidistant from $x=-6\text{.}$

The value of the $y$-coordinate of the $y$-intercept is determined as follows.

\begin{align*} y\amp=-2(\highlight{0}+6)^2+8\\ y\amp=-2 \cdot 36 +8\\ y\amp=-72+8\\ y\amp=-64 \end{align*}

The $y$-intercept of the parabola is $(0,-64)\text{.}$

Example8.4.2

Determine the vertex, axis of symmetry, concavity, and all intercepts for the parabola with equation

\begin{equation*} y=\frac{2}{3}(x-4)^2+1\text{.} \end{equation*}

Solution

The vertex is the point $(4,1)$ and the axis of symmetry is the vertical line with equation $x=4\text{.}$

Because the value of $a\text{,}$ $\frac{2}{3}\text{,}$ is positive, the parabola is concave up.

To determine the $x$-coordinates of the $x$-intercepts, we need to solve the equation $\frac{2}{3}(x-4)^2+1=0\text{.}$ Because of the form of this equation, by far the fastest route to solution is the square root method.

\begin{align*} \frac{2}{3}(x-4)^2+1\amp=0\\ \frac{2}{3}(x-4)^2+1\subtractright{1}\amp=0\subtractright{1}\\ \frac{2}{3}(x-4)^2\amp=-1\\ \multiplyleft{\frac{3}{2}}\frac{2}{3}(x-4)^2\amp=\multiplyleft{\frac{3}{2}}-1\\ (x-4)^2\amp=-\frac{3}{2}\\ x-4\amp=\pm\sqrt{-\frac{3}{2}} \end{align*}

Because $\sqrt{-\frac{3}{2}}$ is not a real number, the parabola has no $x$-intercepts. Note that this is consistent with the fact that the parabola opens upward and the vertex of the parabola is above the $x$-axis.

The value of the $y$-coordinate of the $y$-intercept is determined as follows.

\begin{align*} y\amp=\frac{2}{3}(\highlight{0}-4)^2+1\\ y\amp=\frac{2}{3}(16)-4\\ y\amp=\frac{32}{3}-\frac{12}{3}\\ y\amp=\frac{20}{3} \end{align*}

The $y$-intercept of the parabola is $\left(0,\frac{20}{3}\right)\text{.}$

Example8.4.3

Determine an equation that would produce the parabola shown in Figure 8.4.4. State the equation in both graphing form and standard form.

$The graph of a parabola that opens concave down from the vertex $(-2,6)\text{.}$ The parabola also passes through the point $(2,-6)\text{.}$$

Figure8.4.4Determine an equation for the parabola

Solution

From the vertex of the parabola, $(-2,6)\text{,}$ we know that the graphing form of the equation of the parabola is

\begin{equation*} y=a(x+2)^2+6 \end{equation*}

for some unknown value of $a\text{.}$ We can use the ordered pair $(2,-6)$ to determine the value of $a\text{.}$ Note that the presence of the point $(2,-6)$ on the parabola tells us that the value of $y$ is $-6$ when the value of $x$ is $2\text{.}$

\begin{align*} \highlightr{y}\amp=a(\highlight{x}+2)^2+6\\ \highlightr{-6}\amp=a(\highlight{2}+2)^2+6\\ -6\amp=16a+6\\ -6\subtractright{6}\amp=16a+6\subtractright{6}\\ -12\amp=16a\\ \divideunder{-12}{16}\amp=\divideunder{16a}{16}\\ -\frac{3}{4}\amp=a \end{align*}

So we now know that the graphing form of the the equation for the parabola is

\begin{equation*} y=\frac{3}{4}(x+2)^2+6\text{.} \end{equation*}

We need to expand the right side of the equation to determine the standard form of the equation.

\begin{align*} y\amp=\frac{3}{4}(x+2)^2+6\\ y\amp=\frac{3}{4}(x+2)(x+2)+6\\ y\amp=\frac{3}{4}(x^2+4x+4)+6\\ y\amp=\frac{3}{4}x^2+3x+3+6\\ y\amp=\frac{3}{4}x^2+3x+9 \end{align*}

The standard form of the equation for the parabola is

\begin{equation*} y=\frac{3}{4}x^2+3x+9\text{.} \end{equation*}

Subsection8.4.1Exercises

For each equation, state the vertex, axis of symmetry, concavity, and all intercepts of the parabola.

1

$y=2(x-3)^2-12$

Solution

The equation $y=2(x-3)^2-12$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=2\text{,}$ $h=3\text{,}$ and $k=-12\text{.}$ This tells us that the vertex of the parabola is $(3,-12)$ and that the axis of symmetry is $x=3\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.

\begin{align*} y\amp=2(\highlight{0}-3)^2-12\\ \amp=2(-3)^2-12\\ \amp=6 \end{align*}

So the $y$-intercept is $(0,6)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.

\begin{align*} 2(x-3)^2-12\amp=0\\ 2(x-3)^2-12\addright{12}\amp=0\addright{12}\\ 2(x-3)^2\amp=12\\ \divideunder{2(x-3)^2}{2}\amp=\divideunder{12}{2}\\ (x-3)^2\amp=6\\ x-3\amp=\pm\sqrt{6}\\ x-3\addright{3}\amp=\pm\sqrt{6}\addright{3}\\ x\amp=3\pm\sqrt{6} \end{align*}

So, the $x$-intercepts are $(3-\sqrt{6},0)$ and $(3-\sqrt{6},0)\text{.}$

2

$y=-(x+5)^2$

Solution

The equation $y=-(x+5)^2$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=-1\text{,}$ $h=-5\text{,}$ and $k=0\text{.}$ This tells us that the vertex of the parabola is $(-5,0)$ and that the axis of symmetry is $x=-5\text{.}$ Because $a$ is negative, we also know that the parabola is concave down. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.

\begin{align*} y\amp=-(\highlight{0}+5)^2\\ \amp=-(5)^2\\ \amp=-25 \end{align*}

So the $y$-intercept is $(0,-25)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.

\begin{align*} -(x+5)^2\amp=0\\ x+5\amp=0\\ x+5\subtractright{5}\amp=0\subtractright{5}\\ x\amp=-5 \end{align*}

So, the only $x$-intercept is $(-5,0)\text{.}$

3

$y=-4x^2+3$

Solution

The equation $y=-4x^2+3$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=-4\text{,}$ $h=0\text{,}$ and $k=3\text{.}$ This tells us that the vertex of the parabola is $(0,3)$ and that the axis of symmetry is $x=0$ (i.e. the $y$-axis). Because $a$ is negative, we also know that the parabola is concave down. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.

\begin{align*} y\amp=-4 \cdot (\highlight{0})^2+3\\ \amp=3 \end{align*}

So the $y$-intercept is $(0,3)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.

\begin{align*} -4x^2+3\amp=0\\ -4x^2+3\subtractright{3}\amp=0\subtractright{3}\\ -4x^2\amp=-3\\ \divideunder{-4x^2}{-4}\amp=\divideunder{-3}{-4}\\ x^2\amp=\frac{3}{4}\\ x\amp=\pm\sqrt{\frac{3}{4}}\\ x\amp=\pm\frac{\sqrt{3}}{2} \end{align*}

So, the $x$-intercepts are $\left(-\frac{\sqrt{3}}{2},0\right)$ and $\left(\frac{\sqrt{3}}{2},0\right)\text{.}$

4

$y=3(x-1)^2+4$

Solution

The equation $y=3(x-1)^2+4$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=3\text{,}$ $h=1\text{,}$ and $k=4\text{.}$ This tells us that the vertex of the parabola is $(1,4)$ and that the axis of symmetry is $x=1\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.

\begin{align*} y\amp=3(\highlight{0}-1)^2+4\\ \amp=3(-1)^2+4\\ \amp=7 \end{align*}

So the $y$-intercept is $(0,7)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.

\begin{align*} 3(x-1)^2+4\amp=0\\ 3(x-1)^2+4\subtractright{4}\amp=0\subtractright{4}\\ 3(x-1)^2\amp=-4\\ \divideunder{3(x-1)^2}{3}\amp=\divideunder{-4}{3}\\ (x-1)^2\amp=-\frac{4}{3}\\ x-1=\pm\sqrt{-\frac{4}{3}} \end{align*}

Because the square roots of negative numbers are not real numbers, this parabola has no $x$-intercepts. This is concistent with the fact that the parabola opens upward and that its vertex is already above the $x$-axis.

5

$y=5(x+6)^2-44$

Solution

The equation $y=5(x+6)^2-44$ is already in the graphing form $y=a(x-h)^2+k\text{,}$ so we can see straight away that $a=5\text{,}$ $h=-6\text{,}$ and $k=-44\text{.}$ This tells us that the vertex of the parabola is $(-6,-44)$ and that the axis of symmetry is $x=-6\text{.}$ Because $a$ is positive, we also know that the parabola is concave up. To determine the $y$-intercept, we need to determine the value of $y$ when the value of $x$ is zero.

\begin{align*} y\amp=5(\highlight{0}+6)^2-44\\ \amp=5(6)^2-44\\ \amp=136 \end{align*}

So the $y$-intercept is $(0,136)\text{.}$ To determine the $x$-intercepts, we need to find the points that have a $y$-coordinate of zero.

\begin{align*} 5(x+6)^2-44\amp=0\\ 5(x+6)^2-44\addright{44}\amp=0\addright{44}\\ 5(x+6)^2\amp=44\\ \divideunder{5(x+6)^2}{5}\amp=\divideunder{44}{5}\\ (x+6)^2\amp=\frac{44}{5}\\ x+6\amp=\pm\sqrt{\frac{44}{5}}\\ x+6\amp=\pm\frac{\sqrt{44}}{\sqrt{5}}\\ x+6\amp=\pm\frac{\sqrt{44}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}\\ x+6\amp=\pm\frac{\sqrt{44} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{\sqrt{4 \cdot 11} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{\sqrt{4} \cdot \sqrt{11} \cdot \sqrt{5}}{5}\\ x+6\amp=\pm\frac{2 \cdot \sqrt{11} \sqrt{5}}{5}\\ x+6\amp=\pm\frac{2\sqrt{55}}{5}\\ x+6\subtractright{6}\amp=\pm\frac{2\sqrt{55}}{5}\subtractright{6}\\ x\amp=-6\pm\frac{2\sqrt{55}}{5}\\ x\amp=\frac{-30\pm 2\sqrt{55}}{5} \end{align*}

So, the $x$-intercepts are $\left(\frac{-30-2\sqrt{55}}{5},0\right)$ and $\left(\frac{-30+2\sqrt{55}}{5},0\right)$

For each graph, determine an equation that would produce the parabola. State the equation in both graphing form and standard form.

6

The graph of a parabola that opens concave up from the vertex $(3,-6)\text{.}$ The parabola also passes through the point $(0,-3)\text{.}$ — Figure8.4.5Determine an equation for the parabola

Solution

The vertex of the parabola shown in Figure 8.4.5 is $(3,-6)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x-3)^2-6$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{0},\highlight{-3})$ in that equation to determine the value of $a\text{.}$

\begin{align*} \highlight{-3}\amp=a(\highlightr{0}-3)^2-6\\ -3\amp=9a-6\\ -3\addright{6}\amp=9a-6\addright{6}\\ 3\amp=9a\\ \divideunder{3}{9}\amp=\divideunder{9a}{9}\\ \frac{1}{3}\amp=a \end{align*}

So the graphing form of the equation of the parabola is

\begin{equation*} y=\frac{1}{3}(x-3)^2-6\text{.} \end{equation*}

To get the standard for of the equation, we simply need to expand and simplify the graphing form of the equation.

\begin{align*} y\amp=\frac{1}{3}(x-3)^2-6\\ y\amp=\frac{1}{3}(x-3)(x-3)-6\\ y\amp=\frac{1}{3}(x^2-6x+9)-6\\ y\amp=\frac{1}{3}x^2-2x+3-6\\ y\amp=\frac{1}{3}x^2-2x-3 \end{align*}

So the standard for of the equation is

\begin{equation*} y=\frac{1}{3}x^2-2x-3\text{.} \end{equation*}

7

The graph of a parabola that opens concave down from the vertex $(0,5)\text{.}$ The parabola also passes through the point $(2,1)\text{.}$ — Figure8.4.6Determine an equation for the parabola

Solution

The vertex of the parabola shown in Figure 8.4.6 is $(0,5)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x-0)^2+5$ or merely $y=ax^2+5$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{2},\highlight{1})$ in that equation to determine the value of $a\text{.}$

\begin{align*} \highlight{1}\amp=(\highlightr{2})^2+5\\ 1\amp=4a+5\\ 1\subtractright{5}\amp=4a+5\subtractright{5}\\ -4\amp=4a\\ \divideunder{-4}{4}\amp=\divideunder{4a}{4}\\ -1\amp=a \end{align*}

So the graphing form of the equation of the parabola is $y=-1 \cdot x^2+5$ or simply

\begin{equation*} y=-x^2+5\text{.} \end{equation*}

We'll note that the last stated equation is also the standard form of the equation of the parabola.

8

The graph of a parabola that opens concave up from the point $(-2,-1)\text{.}$ The graph also passes through the point $(0,3)\text{.}$ — Figure8.4.7Determine an equation for the parabola

Solution

The vertex of the parabola shown in Figure 8.4.7 is $(-2,-1)\text{.}$ This tells us that the graphing form of the equation has form $y=a(x+2)^2-1$ for some undetermined value of $a\text{.}$ We can use the ordered pair $(\highlightr{0},\highlight{3})$ in that equation to determine the value of $a\text{.}$

\begin{align*} \highlight{3}\amp=a(\highlightr{0}+2)^2-1\\ 3\amp=4a-1\\ 3\addright{1}\amp=4a-1\addright{1}\\ 4\amp=4a\\ \divideunder{4}{4}\amp=\divideunder{4a}{4}\\ 1\amp=a \end{align*}

So the graphing form of the equation of the parabola is $y=1 \cdot (x+2)^2-1$ or simply

\begin{equation*} y=(x+2)^2-1\text{.} \end{equation*}

To get the standard for of the equation, we simply need to expand and simplify the graphing form of the equation.

\begin{align*} y\amp=(x+2)^2-1\\ y\amp=(x+2)(x+2)-1\\ y\amp=x^2+4x+4-1\\ y\amp=x^2+4x+3 \end{align*}

So the standard for of the equation is

\begin{equation*} y=x^2+4x+3\text{.} \end{equation*}