Answer is: option1

Solution:

We are given:

1. The graph is of \( f'(x) \), the derivative of \( f(x) \), and it's a straight line.
2. The line has x-intercept at \( x = -1.5 \) and y-intercept at \( y = 3 \), so we can find the equation of \( f'(x) \).
3. We're told that \( f(3) = 11 \), and we're asked to find \( f(-3) \).

We can use the two points on the line:

1. Point A: \( (-1.5, 0) \)
2. Point B: \( (0, 3) \)

Slope \( m = \frac{3 - 0}{0 - (-1.5)} = \frac{3}{1.5} = 2 \)

Using point-slope form:

\[ y - 3 = 2(x - 0) \Rightarrow y = 2x + 3 \]

So,

\[ f'(x) = 2x + 3 \]

\[ f(x) = \int f'(x)\, dx = \int (2x + 3)\, dx = x^2 + 3x + C \]

Use the given value \( f(3) = 11 \) to solve for \( C \)

\[ f(3) = (3)^2 + 3(3) + C = 9 + 9 + C = 18 + C \]

\[ 11 = 18 + C \Rightarrow C = -7 \]

\[ f(x) = x^2 + 3x - 7 \]

Find \( f(-3) \)

\[ f(-3) = (-3)^2 + 3(-3) - 7 = 9 - 9 - 7 = -7 \]

Final Answer: \( f(-3) = -7 \)