Chapter 9.2, Problem 3BGP

To determine

To draw: The graph of the equation $y=x^2+6x-4$ .

Answer to Problem 3BGP

The graph of the equation is shown in figure (1).

Explanation of Solution

Given:

The given equation is $y=x^2+6x-4$ .

Calculation:

Rewrite the given quadratic equation in the standard form by completing the square.

  $\begin{array}{c}y=\left(x^2+6x+9\right)-4-9\\ y={\left(x+3\right)}^2-13\end{array}$

Compare the equation $y={\left(x+3\right)}^2-13$ with the standard equation $y=a{\left(x-h\right)}^2+k$ .

  $\begin{array}{c}a=1\\ h=-3\\ k=-13\end{array}$

Therefore, the vertex of the given equation is $\left(-3,-13\right)$ .

Axis of symmetry will be given shown below.

  $\begin{array}{c}x=h\\ =-3\end{array}$

Therefore, the axis of symmetry is $x=-3$ .

Find the focus.

  $\begin{array}{c}\text{Focus}=\left(-13,-3+\frac{1}{4\left(1\right)}\right)\\ =\left(-13,-\frac{11}{4}\right)\end{array}$

Find the directrix.

  $\begin{array}{c}y=k-\frac{1}{4a}\\ =-13-\frac{1}{4}\\ =-\frac{53}{4}\end{array}$

As it is found that the value of $a=1$ is negative so by definition the parabola opens towards right.

The length of the latus rectum is $\left|\frac{1}{\left(1\right)}\right|=1$ unit.

Draw the graph of the equation $y={\left(x+3\right)}^2-13$ as follows.

  

Figure (1)

Therefore, the graph of the equation is shown in figure (1).