Chapter 2.1, Problem 35E

Solution

Describe the graph of the function $f\left(x\right)=10-16x-x^2$ using the completing square method. Support your answer graphically.

The given parabola is a downward opening parabola with vertex $\left(-8,74\right)$ and the axis of symmetry is $x=-8$ . The y -intercept of the graph is given by $\left(0,10\right)$ .

Given:

  $f\left(x\right)=10-16x-x^2$ .

Concept Used:

• The vertex form of a parabola is given by $y=a{(x-h)}^2+k$
• The axis of the graph of the parabola is $x=h$

Calculation:

The given function is $f\left(x\right)=10-16x-x^2$ .

Use completing square method to write to write the function in vertex form

  $\begin{array}{c}f\left(x\right)=10-16x-x^2\\ =-x^2-16x+10\\ =-\left(x^2+16x-10\right)\\ =-\left(x^2+16x-10+8^2-8^2\right)\left(\text{Add and subtract }{\left(8\right)}^2\right)\\ =-\left({\left(x+8\right)}^2-10-8^2\right)\\ =-{\left(x+8\right)}^2+74\end{array}$

So, the vertex form of the parabola is $f\left(x\right)=-{\left(x+8\right)}^2+74$

Comparing this function with the vertex form $y=a{(x-h)}^2+k$

  $h=-8,k=74,a=-1$

Hence, the vertex of the parabola is given by

  $\begin{array}{l}\left(h,k\right)\\ =\left(-8,74\right)\end{array}$

And the axis of the parabola is given by

  $\begin{array}{c}x=h\\ x=-8\end{array}$

For y-intercept of the parabola

  $\begin{array}{l}y=10-16\left(0\right)-{\left(0\right)}^2\\ y=10\end{array}$

Thus, the y -intercept of the graph is given by $\left(0,10\right)$

And the value for a is $-1$ , which is less than zero. Hence, the parabola opens downward.

Therefore, the given parabola is a downward opening parabola with vertex $\left(-8,74\right)$ and the axis of symmetry is $x=-8$ . The y -intercept of the graph is given by $\left(0,10\right)$ .

Below is the graph of the given function where we can see the vertex, axis of symmetry and y-intercept.