Chapter 7, Problem 1RE

For Exercises 1– 6, sketch the graph of each parabola. Determine the x- and y-intercepts, vertex, axis of symmetry, focus, and directrix for each.

$y=x^2+4x-12$

To determine

To graph: The parabola $y=x^2+4x-12$, and also determine the x-and y-intercepts, vertex, axis of symmetry, focus and directrix of parabola.

Explanation of Solution

Given:

The equation is $y=x^2+4x-12$.

Graph:

To plot the graph of the function, consider the below mentioned steps and use the Ti-83 calculator:

The standard form of parabola ${\left(x-h\right)}^2=4a\left(y-k\right)$, where $\left(h,k\right)$ is the vertex of the parabola.

Step 1:

Press the [Y=] key to enter the equations for $y$.

Step 2:

Enter the equations in ${\text{Y}}_{\text{1}}$. Here, ${\text{Y}}_{\text{1}}=x^2+4x-12$.

Step 3:

Press [WINDOW] key to set the window and then edit the values as follows:

$\begin{array}{ccc}\hfill X\min & =& -20\hfill \\ \hfill X\max & =& 20\hfill \\ \hfill Xscale& =& 1\hfill \end{array}$

$\begin{array}{ccc}\hfill Y\min & =& -20\hfill \\ \hfill Y\max & =& 20\hfill \\ \hfill Yscale& =& 1\hfill \end{array}$

Step 4:

Press the [GRAPH] key to plot the graph.

Step 5:

Press the [TRACE] key to locate the intercept.

The graph thus obtained as follows:

The equation of the parabola $y=x^2+4x-12$ can be rewritten in the form of $y=a{\left(x-h\right)}^2+k$ as shown below:

$\begin{array}{ccc}\hfill y& =& x^2+4x-12\hfill \\ \hfill y& =& 1\left(x^2+4x\right)-12\hfill \\ \hfill y& =& 1\left(x^2+4x+4-4\right)-12\hfill \\ \hfill y& =& {\left(x+2\right)}^2-16\hfill \end{array}$

The y-intercept of the equation can be computed by putting the value of x equal to 0 as shown below:

$y={\left(0+2\right)}^2-16=-12$

Therefore, the y-intercept is $\left(0,-12\right)$.

The x-intercept of the equation can be computed by putting the value of y equal to 0 as shown below:

$\begin{array}{ccc}\hfill {\left(x+2\right)}^2-16& =& 0\hfill \\ \hfill x+2& =& \sqrt{16}\hfill \\ \hfill x+2& =& \pm 4\hfill \\ \hfill x& =& 2,-6\hfill \end{array}$

Therefore, the x-intercepts are $\left(2,0\right),\left(-6,0\right)$.

The vertex in the equation of the parabola $y=a{\left(x-h\right)}^2+k$ is $\left(h,k\right)$.

Therefore the vertex of the equation is $\left(-2,-16\right)$.

The focus of the equation can be computed as $\left(h,k+p\right)$ and directrix can be given as $y=k-p$

Since, in the equation $y={\left(x+2\right)}^2-16$, $a=1$, the focus can be computed as shown below:

$\begin{array}{ccc}\hfill a& =& \frac{1}{4p}\hfill \\ \hfill 1& =& \frac{1}{4p}\hfill \\ \hfill p& =& \frac{1}{4}\hfill \end{array}$

Therefore, focus can be written as $\left(-2,-16+\frac{1}{4}\right)=\left(-2,\frac{-63}{4}\right)$ and

The directrix is $y=-16-\frac{1}{4}$

$y=\frac{-65}{4}$

The axis of the symmetry for the equation $y={\left(x+2\right)}^2-16$ is the vertical line $x=-2$.

Hence, the x-intercepts are $\left(2,0\right),\left(-6,0\right)$, y-intercept is $\left(0,-12\right)$, vertex is $\left(-2,-16\right)$, focus is $\left(-2,\frac{-63}{4}\right)$, directrix is $y=\frac{-65}{4}$, and axis of symmetry is $x=-2$.

Interpretation:

The x-intercepts are $\left(2,0\right),\left(-6,0\right)$, y-intercept is $\left(0,-12\right)$, vertex is $\left(-2,-16\right)$, focus is $\left(-2,\frac{-63}{4}\right)$, directrix is $y=\frac{-65}{4}$, and axis of symmetry is $x=-2$.