Find the slope of the curve at the given point P and an equation of the tangent line at P.
-$y = x ^ { 2 } + 5 x , P ( 4,36 )$
A) slope is $\frac { 1 } { 20 } ; y = \frac { x } { 20 } + \frac { 1 } { 5 }$
B) slope is $- \frac { 4 } { 25 } ; y = - \frac { 4 x } { 25 } + \frac { 8 } { 5 }$
C) slope is $- 39 ; y = - 39 x - 80$
D) slope is $13 ; y = 13 x - 16$

Question

Quizplus · Accepted Answer

To find the slope of the curve at point P, we first differentiate $y = x^2 + 5x$ with respect to $x$, which gives us $y' = 2x + 5$. Substituting $x = 4$ into $y'$, we get $2(4) + 5 = 13$, which is the slope at point P. To find the equation of the tangent line at P, we use the point-slope form $y - y_1 = m(x - x_1)$, where $m = 13$ and $P(x_1, y_1) = (4, 36)$. Substituting these values, we get $y - 36 = 13(x - 4)$, which simplifies to $y = 13x - 52 + 36$, or $y = 13x - 16$.

Find the Slope of the Curve at the Given Point y=x2+5x,P(4,36)y = x ^ { 2 } + 5 x , P ( 4,36 )y=x2+5x,P(4,36)

Find the Slope of the Curve at the Given Point $y = x ^ { 2 } + 5 x , P ( 4,36 )$