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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.

plain text
Figure8.4.4Determine equations 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

plain text
Figure8.4.5Determine equations 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

plain text
Figure8.4.6Determine equations 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

plain text
Figure8.4.7Determine equations 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*}