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 \(f(-2) = \)?
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 \(f(-2) = \)?
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Integrate the slope function to retrieve the original function, ensuring to apply any given initial conditions correctly.
Updated On: Mar 11, 2025
- \(0\)
- \(4\)
- \(-6\)
- \(-13\)
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The Correct Option is B
Solution and Explanation
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, \(f(-2) = 4\).
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