Question

In the following exercises, ⓐ solve graphically and ⓑ write the solution in interval notation$-x^2+2x+24<0$

Short answer

The solution in interval notation is $\left(-\infty \hspace{0.17em},\hspace{0.17em}-4\right)\cup \left(6,\hspace{0.17em}\infty \hspace{0.17em}\right)$

Step-by-step solution

Step 1. Given information

$-x^2+2x+24<0$

Step 2. Solve graphically   

The quadratic equation is in standard form.

$-x^2+2x+24<0$

Graph the function $f\left(x\right)=-x^2+2x+24$

The parabola opens downward.

Find the line of symmetry

$$\begin{gathered} x=-\frac{b}{2a} \\ x=1 \end{gathered}$$

Find the vertex

$$\begin{gathered} f\left(x\right)=-x^2+2x+24 \\ f\left(1\right)=-{\left(1\right)}^2+2\left(1\right)+24 \\ f\left(1\right)=25 \end{gathered}$$

vertex is $\left(1,25\right)$

Find the x-intercepts. Let $f\left(x\right)=0$

$$\begin{gathered} f\left(x\right)=-x^2+2x+24 \\ 0=-\left(x+4\right)\left(x-6\right) \\ x=-4orx=6 \end{gathered}$$

The x-intercepts are $\left(-4,0\right)and\left(6,0\right)$

The graph of the parabola is

Step 3. Solution

The solution in interval notation is $\left(-\infty \hspace{0.17em},\hspace{0.17em}-4\right)\cup \left(6,\hspace{0.17em}\infty \hspace{0.17em}\right)$