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.
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.
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
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*}
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.
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.
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.
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 \(2\text{.}\)
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.
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.
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.
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.
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.
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.
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 \(a\text{.}\)
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 \(a\text{.}\)
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 \(a\text{.}\)
