Find the equation of the tangent line to the curve when $\mathbf{x}$ has the given value.
-$f(x)=x^{2}+11 x-15 ; x=1$
A) $y=13 x-16$
B) $y=-\frac{4 x}{25}+\frac{8}{5}$
C) $y=-39 x-80$
D) $y=\frac{x}{20}+\frac{1}{5}$

Question

Quizplus · Accepted Answer

The derivative of $f(x) = x^2 + 11x - 15$ is $f'(x) = 2x + 11$. At $x = 1$, the slope is $f'(1) = 2(1) + 11 = 13$. The point on the curve is $(1, f(1)) = (1, -3)$. The equation of the tangent line is $y - (-3) = 13(x - 1)$, which simplifies to $y = 13x - 16$.

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