Question

(u5) the graph of the function f, defined by $f(x) = \frac{1}{3}(x + 2)^2$, is shown in the xy-plane below. if the function g (not shown) is defined by $g(x) = -x - 2$, what are the possible values of x such that $f(x)=g(x)$?
$x=\square$ & $x=\square$ (order does not matter)

Explanation:

Step1: Set f(x) equal to g(x)

We have \( f(x) = \frac{1}{3}(x + 2)^2 \) and \( g(x) = -x - 2 \). So, set them equal:
\( \frac{1}{3}(x + 2)^2 = -x - 2 \)

Step2: Eliminate the fraction

Multiply both sides by 3 to get rid of the fraction:
\( (x + 2)^2 = -3x - 6 \)

Step3: Expand the left side

Expand \( (x + 2)^2 \) using the formula \( (a + b)^2 = a^2 + 2ab + b^2 \):
\( x^2 + 4x + 4 = -3x - 6 \)

Step4: Move all terms to one side

Bring all terms to the left side to form a quadratic equation:
\( x^2 + 4x + 4 + 3x + 6 = 0 \)
Simplify:
\( x^2 + 7x + 10 = 0 \)

Step5: Factor the quadratic

Factor \( x^2 + 7x + 10 \). We need two numbers that multiply to 10 and add to 7. Those numbers are 2 and 5:
\( (x + 2)(x + 5) = 0 \)

Step6: Solve for x

Set each factor equal to zero:
\( x + 2 = 0 \) or \( x + 5 = 0 \)
Solving these gives \( x = -2 \) or \( x = -5 \)

Answer:

\( x = -5 \) & \( x = -2 \)