Question

If the slope of the tangent drawn at any point \((x, y)\) on the curve \(y = f(x)\) is \(y = 6x^2 + 10x - 9\) and \(f(2) = 0\), then \(f(-2) = \)?

• A. \(0\)
• B. \(4\)
• C. \(-6\)
• D. \(-13\)

Hint:

Integrate the slope function to retrieve the original function, ensuring to apply any given initial conditions correctly.

Answer 1

The Correct Option is B

Given the slope \(y'\) of the tangent is \(y' = 6x^2 + 10x - 9\) and \(f(2) = 0\). We integrate the derivative to find \(f(x)\): \[ f(x) = \int (6x^2 + 10x - 9) \, dx = 2x^3 + 5x^2 - 9x + C \] Using \(f(2) = 0\): \[ f(2) = 2(2)^3 + 5(2)^2 - 9(2) + C = 16 + 20 - 18 + C = 18 + C = 0 \] \[ C = -18 \] Therefore, the function is: \[ f(x) = 2x^3 + 5x^2 - 9x - 18 \] Calculating \(f(-2)\): \[ f(-2) = 2(-2)^3 + 5(-2)^2 - 9(-2) - 18 = -16 + 20 + 18 - 18 = 4 \] Thus, \(f(-2) = 4\).