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

Question

Quizplus · Accepted Answer

To find the slope of the curve at point P(1,3), we first differentiate the given function $y = x^3 - 2x^2 + 4$ with respect to $x$, which gives us $y' = 3x^2 - 4x$. Substituting $x = 1$ into $y'$, we get $3(1)^2 - 4(1) = 3 - 4 = -1$, so the slope at P is $-1$. To find the equation of the tangent line at P, we use the point-slope form $y - y_1 = m(x - x_1)$, where $m = -1$, $x_1 = 1$, and $y_1 = 3$. Substituting these values, we get $y - 3 = -1(x - 1)$, which simplifies to $y = -x + 4$.

Find the Slope of the Curve at the Given Point y=x3−2x2+4,P(1,3)y = x ^ { 3 } - 2 x ^ { 2 } + 4 , P ( 1,3 )y=x3−2x2+4,P(1,3)

Find the Slope of the Curve at the Given Point $y = x ^ { 3 } - 2 x ^ { 2 } + 4 , P ( 1,3 )$