Find the slope of the curve at the given point P and an equation of the tangent line at P.
-$$y = x ^ { 3 } - 2 x ^ { 2 } + 4 , P ( 3,13 )$$
A) slope is $1 ; y = x - 32$
B) slope is $15 ; y = 15 x + 13$
C) slope is $15 ; y = 15 x - 32$
D) slope is $0 ; \mathrm { y } = - 32$

Question

Quizplus · Accepted Answer

To find the slope of the curve at point P, we first differentiate the given function with respect to x: $$y' = 3x^2 - 4x.$$ Substituting $x = 3$ into the derivative gives the slope at P: $$y'(3) = 3(3)^2 - 4(3) = 27 - 12 = 15.$$ To find the equation of the tangent line at P, we use the point-slope form of a line: $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point P. Substituting $m = 15$ and $P(3, 13)$ gives: $$y - 13 = 15(x - 3).$$ Simplifying, we get: $$y = 15x - 45 + 13 = 15x - 32.$$

Find the Slope of the Curve at the Given Point y=x3−2x2+4,P(3,13)y = x ^ { 3 } - 2 x ^ { 2 } + 4 , P ( 3,13 )y=x3−2x2+4,P(3,13)

Find the Slope of the Curve at the Given Point $y = x ^ { 3 } - 2 x ^ { 2 } + 4 , P ( 3,13 )$