Horizontal Parabolas-2

If we subtract LaTeX: k $k$ from each side of the equation for a vertical parabola, $LaTeX: y=a\left(x-h\right)^2+k\:$ $y=a\left(x-h\right)^2+k\:$ , and interchange the roles of LaTeX: x-h $x-h$ and LaTeX: y-k, $y-k,$ the new equation also has a parabola as its graph.

Demo of subtracting k and interchanging x-h and y-k.PNG

While the graph of $LaTeX: y=a\left(x-h\right)^2+k\:$ $y=a\left(x-h\right)^2+k\:$ has a vertical axis of symmetry, the graph of $LaTeX: x-h=a\left(y-k\right)^2\:$ $x-h=a\left(y-k\right)^2\:$ has a horizontal axis of symmetry. The graph of $LaTeX: y=a\left(x-h\right)^2+k\:$ $y=a\left(x-h\right)^2+k\:$ is the graph of a function (specifically a quadratic function), while the graph of the second equation is not. Its graph fails the vertical line test.

Parabola with a Horizontal Axis of Symmetry

The parabola with vertex $LaTeX: \left(h,k\right)$ $\left(h,k\right)$ and the horizontal line LaTeX: y=k $y=k$ as axis of symmetry has an equation of the following form.

$LaTeX: x-h=a\left(y-k\right)^2\:$ $x-h=a\left(y-k\right)^2\:$

The parabola opens to the right if LaTeX: a>0 $a>0$ and to the left if LaTeX: a<0 $a<0$ .

Note:

When the vertex (h, k) is (0, 0) and a = 1 in $LaTeX: y=a\left(x-h\right)^2+k\:$ $y=a\left(x-h\right)^2+k\:$ and $LaTeX: x-h=a\left(y-k\right)^2\:$ $x-h=a\left(y-k\right)^2\:$ the equations LaTeX: y=x^2 $y=x^2$ and LaTeX: x=y^2 $x=y^2$ , respectively, result. The graphs are mirror images of each other with respect to the line $y=x$ .

Coordinate Axes with both of these graphs and the y=x line shown.PNG

Example

Graph $LaTeX: x+3=\left(y-2\right)^2$ $x+3=\left(y-2\right)^2$ .

Give the domain and range.

Solution

The graph of $LaTeX: x+3=\left(y-2\right)^2$ $x+3=\left(y-2\right)^2$ , or $LaTeX: x-\left(-3\right)=\left(y-2\right)^2$ $x-\left(-3\right)=\left(y-2\right)^2$ , has vertex (–3, 2) and opens to the right because a = 1, and 1 > 0. Plotting a few additional points gives the graph shown.

Point coordinates and a graph of the parabola.PNG

Note that the graph is symmetric about its axis,
y = 2. The domain is $LaTeX: [-3,\infty )$ $[-3,\infty )$ , and the range is
$LaTeX: \left(-\infty,\infty\right)$ $\left(-\infty,\infty\right)$ .

Example

Graph LaTeX: x=2y^2+6y+5 $x=2y^2+6y+5$ .

Give the domain and range.

Solution

First we need to complete the square to get the equation into the form we need.

LaTeX: x=2y^2+6y+5 $x=2y^2+6y+5$

Factor out the coefficient of the y squared term, 2.

$LaTeX: x=2\left(y^2+3y\:\:\:\right)+5$ $x=2\left(y^2+3y\:\:\:\right)+5$

Complete the square by adding and subtracting $LaTeX: \left[\frac{1}{2}\cdot3\right]^2=\frac{9}{4}$ $\left[\frac{1}{2}\cdot3\right]^2=\frac{9}{4}$ .

$LaTeX: x=2\left(y^2+3y+\frac{9}{4}-\frac{9}{4}\right)+5$ $x=2\left(y^2+3y+\frac{9}{4}-\frac{9}{4}\right)+5$

Using the distributive property and simplifying gives us:

$LaTeX: x=2\left(y^2+3y+\frac{9}{4}\right)+2\left(-\frac{9}{4}\right)+5$ $x=2\left(y^2+3y+\frac{9}{4}\right)+2\left(-\frac{9}{4}\right)+5$

$LaTeX: x=2\left(y+\frac{3}{2}\right)^2+\frac{1}{2}$ $x=2\left(y+\frac{3}{2}\right)^2+\frac{1}{2}$

$LaTeX: x-\frac{1}{2}=2\left(y+\frac{3}{2}\right)^2$ $x-\frac{1}{2}=2\left(y+\frac{3}{2}\right)^2$

The vertex of the parabola is $LaTeX: \left(\frac{1}{2},-\frac{3}{2}\right)$ $\left(\frac{1}{2},-\frac{3}{2}\right)$ . The axis is the horizontal line y = –3/2. Using the vertex and the axis and plotting a few additional points gives the graph. Let y = 0 to find that the x-intercept is (5,0), and because of symmetry, the point (5,–3) also lies on the graph. The domain is $LaTeX: [\frac{1}{2} ,\infty )$ $[\frac{1}{2} ,\infty )$ and the range is $LaTeX: \left(-\infty,\infty\right)$ $\left(-\infty,\infty\right)$ .