Find the slope of the curve at the given point P and an equation of the tangent line at P.
-$y = x ^ { 2 } + 11 x - 15 , P ( 1 , - 3 )$
A) slope is $- \frac { 4 } { 25 } ; \mathrm { y } = - \frac { 4 \mathrm { x } } { 25 } + \frac { 8 } { 5 }$
B) slope is $- 39 ; y = - 39 x - 80$
C) slope is $\frac { 1 } { 20 } ; y = \frac { x } { 20 } + \frac { 1 } { 5 }$
D) slope is $13 ; y = 13 x - 16$

Question

Quizplus · Accepted Answer

To find the slope of the curve at a given point, we differentiate the given function with respect to $x$. The derivative of $y = x^2 + 11x - 15$ is $y' = 2x + 11$. Substituting $x = 1$ into $y'$ gives $2(1) + 11 = 13$, which is the slope at point $1, -3$. To find the equation of the tangent line at $P(1, -3)$, we use the point-slope form: $y - y_1 = m(x - x_1)$, where $m = 13$, $x_1 = 1$, and $y_1 = -3$. Substituting these values gives $y + 3 = 13(x - 1)$, or $y = 13x - 16$.

Find the Slope of the Curve at the Given Point y=x2+11x−15,P(1,−3)y = x ^ { 2 } + 11 x - 15 , P ( 1 , - 3 )y=x2+11x−15,P(1,−3)

Find the Slope of the Curve at the Given Point $y = x ^ { 2 } + 11 x - 15 , P ( 1 , - 3 )$