Chapter 2.1, Problem 28E

To determine

Tograph: Graph the quadratic function $f\left(x\right)=-\left(x^2+x-30\right)$ and identify vertex, axis of symmetry and x-intercepts.

Explanation of Solution

Given information: we have the given quadratic function is $f\left(x\right)=-\left(x^2+x-30\right)$

Graph:

Using graphing calculator, the graph of the function is shown below

  

Interpretation:

We know that the standard for of a quadratic function $f\left(x\right)=a{\left(x-h\right)}^2+k,a\ne 0$ whose axis of symmetry is $x=h$ and vertex $\left(h,k\right)$ .

For, the above form if $a>0$ then the graph is open upward and when $a<0$ the graph is open downward.

We have the given function is $f\left(x\right)=-\left(x^2+x-30\right)$

It can further written as

  $\begin{array}{c}f\left(x\right)=-\left(x^2+x-30\right)\\ =-{\left(x+\frac{1}{2}\right)}^2+\frac{121}{4}\end{array}$

Comparing with above we get

  $\begin{array}{l}a=-1<0\\ h=\frac{1}{2}\\ k=\frac{121}{4}\end{array}$

Therefore, the vertex of the parabola is $\left(\frac{-1}{2},\frac{121}{4}\right)$ axis of symmetry is $x=\frac{-1}{2}$ and the graph is open downward.

To get the x-intercept we put $f\left(x\right)=0$ and solve for x which gives

  $\begin{array}{l}-\left(x^2+2x-3\right)=0\\ x=5,6\end{array}$