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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*}