Find the equation of the tangent line to the graph of f at the given point.
-$f ( x ) = x ^ { 2 } + 11 x - 15$ at $( 1 , - 3 )$
A) $y = 13 x - 16$
B) $y = 2 x + 11$
C) $y = 11 x$
D) $y = 11 x + 15$

Question

Quizplus · Accepted Answer

The slope of the tangent line at a point on a curve is given by the derivative of the function at that point. The derivative of $f(x) = x^2 + 11x - 15$ is $f'(x) = 2x + 11$. At $x = 1$, $f'(1) = 2(1) + 11 = 13$, so the slope of the tangent line is 13. Using the point-slope form of a line, $y - y_1 = m(x - x_1)$, with $m = 13$ and $(x_1, y_1) = (1, -3)$, we get $y - (-3) = 13(x - 1)$, which simplifies to $y = 13x - 16$.

Find the Equation of the Tangent Line to the Graph f(x)=x2+11x−15f ( x ) = x ^ { 2 } + 11 x - 15f(x)=x2+11x−15

Find the Equation of the Tangent Line to the Graph $f ( x ) = x ^ { 2 } + 11 x - 15$