Question

find the zeros of the function. enter the solutions from least to greatest. f(x) = (x + 2)^2 - 64 lesser x = greater x = show calculator

Explanation:

Step1: Set \( f(x) = 0 \)

To find the zeros of the function \( f(x)=(x + 2)^2-64 \), we set \( f(x)=0 \). So we have the equation:
\[
(x + 2)^2-64 = 0
\]

Step2: Solve for \( x \)

First, we can rewrite the equation as \((x + 2)^2=64\). Then, take the square root of both sides. Remember that if \( y^2 = a\) (\(a\geq0\)), then \(y=\pm\sqrt{a}\). So for \((x + 2)^2 = 64\), we get:
\[
x + 2=\pm\sqrt{64}
\]
Since \(\sqrt{64} = 8\), this simplifies to:
\[
x + 2=\pm8
\]
Now we solve for \(x\) in two cases:

Case 1: \(x + 2 = 8\)

Subtract 2 from both sides:
\[
x=8 - 2=6
\]

Case 2: \(x + 2=-8\)

Subtract 2 from both sides:
\[
x=-8 - 2=-10
\]

Answer:

lesser \( x=-10\)
greater \( x = 6\)