Question

Need help finding the x-intercepts for equation in picture. I can see them on the graph but I need to work it out by solving.

Answer 1

Answer:

The x-intercepts of the function are;

$\begin{gathered} x=-2 \\ \text{and} \\ x=-4 \end{gathered}$

Explanation:

Given the function;

$f(x)=-2(x+3)^2+2$

We want to derive the x-intercepts of the function.

The x-intercept is at f(x)=0;

$\begin{gathered} f(x)=-2(x+3)^2+2=0 \\ -2(x+3)^2+2=0 \\ -2(x^2+6x+9)^{}+2=0 \\ -2x^2-12x-18^{}+2=0 \\ -2x^2-12x-16=0 \\ -x^2-6x-8=0 \\ x^2+6x+8=0 \end{gathered}$

solving for x;

$\begin{gathered} x^2+6x+8=0 \\ x^2+2x+4x+8=0 \\ (x+2)(x+4)=0 \\ x+2=0 \\ x=-2 \\ \text{and} \\ x+4=0 \\ x=-4 \end{gathered}$

Therefore, the x-intercepts of the function are;

$\begin{gathered} x=-2 \\ \text{and} \\ x=-4 \end{gathered}$

Method 2: quadratic root property;

$\begin{gathered} f(x)=-2(x+3)^2+2=0 \\ -2(x+3)^2+2=0 \\ -2(x+3)^2=-2 \\ \text{divide both sides by -2;} \\ (x+3)^2=1 \\ \text{square root both sides;} \\ \sqrt{(x+3)^2}=\sqrt{1} \\ x+3=\pm1 \\ x=-3\pm1 \\ so\text{ the values of x are;} \\ x=-3+1=-2 \\ \text{and} \\ x=-3-1=-4 \end{gathered}$

Therefore, the x-intercepts are;

$\begin{gathered} x=-2 \\ \text{and } \\ x=-4 \end{gathered}$