Chapter 8.3, Problem 15E

To determine

To graph the function:

  $y=-8x^2-16x-9$

And also define it’s domain and range.

Answer to Problem 15E

  

Domain is: $x\in \left(-\infty ,\infty \right)$ , and range is: $y\le -1$ .

Explanation of Solution

Given information:

  $y=-8x^2-16x-9$

Concept Used:

Graph the function $f\left(x\right)=a{x}^2+bx+c$

• The graph opens up when $a>0$ , and the graph opens down when $a<0$ .
• The y -intercept is c.
• The x -coordinate of the vertex of graph is $-\frac{b}{2a}$ .
• The axis of symmetry is $x=-\frac{b}{2a}$ .

Calculation:

In order to graph the function:

  $y=-8x^2-16x-9$

First find the axis/line of symmetry of graph. Axis of symmetry of a function $f\left(x\right)=a{x}^2+bx+c$ is given by:

  $x=-\frac{b}{2a}$

Here,

  $a=-8,b=-16,\text{ and }c=-9$

Thus, axis of symmetry of graph of given function is:

  $\begin{array}{c}x=-\frac{b}{2a}\\ =-\frac{-16}{2\left(-8\right)}\\ =-\frac{-16}{-18}\\ =-1\\ \Rightarrow x=-1\end{array}$

Now, find x -coordinate of the vertex of graph of this function, that is $-\frac{b}{2a}=-1$

And y -coordinate of vertex is found by putting $x=-1$ in equation of function, as:

  $\begin{array}{c}{y|}_{-1}=-8{\left(-1\right)}^2-16\left(-1\right)-9\\ =-8+16-9\\ =-17+16\\ =-1\end{array}$

Thus, vertex of graph of given function is: $\left(-1,-1\right)$ .

Now, find y-intercept to find two more points on the graph. Since y -intercept of the function $f\left(x\right)=a{x}^2+bx+c$ is given by c .

And, $c=-9$ . Thus, y -intercept is $\left(0,-9\right)$ .

Now, since $a=-8<0$ , thus curve of this is open upward.

From the above information plot the vertex, $\left(-1,-1\right)$ , and y -intercept $\left(0,-9\right)$ and make a smooth curve through these points that is open downward and use the axis of symmetry $x=-1$to draw other half of the graph. Using a graphing utility this can be represented as:

  

Now, from the above graph it can be concluded that the domain of function is real number, that is $x\in \left(-\infty ,\infty \right)$ , but the range is every number less than and equals to $-1$ , thus range is $y\le -1$ .