Find the slope of the normal line to the curve $y = 4 x ^ { 2 } - 5 x$ at the point (1, $- 1 )$ .
A)8

B) $- 5$
C)3
D)1
E) $- \frac { 1 } { 3 }$
F) $\frac { 1 } { 5 }$
G) $- \frac { 1 } { 8 }$
H)0

Question

Find the slope of the normal line to the curve $y = 4 x ^ { 2 } - 5 x$ at the point (1, $- 1 )$ .
A)8
 
B) $- 5$
C)3
D)1
E) $- \frac { 1 } { 3 }$
F) $\frac { 1 } { 5 }$
G) $- \frac { 1 } { 8 }$ 
H)0

Quizplus · Accepted Answer

The slope of the tangent line to the curve at a given point is found by taking the derivative of the function and evaluating it at that point. The derivative of $y = 4x^2 - 5x$ is $y' = 8x - 5$. Evaluating this at $x = 1$ gives $8(1) - 5 = 3$, which is the slope of the tangent line. The slope of the normal line, which is perpendicular to the tangent, is the negative reciprocal of the tangent's slope. Therefore, the slope of the normal line is $-\frac{1}{3}$.

Find the Slope of the Normal Line to the Curve y=4x2−5xy = 4 x ^ { 2 } - 5 xy=4x2−5x

Find the Slope of the Normal Line to the Curve $y = 4 x ^ { 2 } - 5 x$