The tangent line to the curve y = $\frac { 1 } { 6 }$ $x ^ { 3 }$ - $\frac { 1 } { 4 }$ $x ^ { 2 }$ - x + 4 is perpendicular to the line $18 x + 9 y = - 37$ at two points on the curve. Find the two points.
A) (2, 0), (-1, 0)
B) $\left( 2 , \frac { 7 } { 3 } \right)$ , $\left. - 1 , \frac { 55 } { 12 } \right)$
C) $\left( 2 , \frac { 7 } { 3 } \right)$ , $\left.( 1 , \frac { 55 } { 12 } \right)$
D) $\left( - 2 , - \frac { 1 } { 3 } \right)$ , $\left( 1 , \frac { 59 } { 12 } \right)$

Question

Quizplus · Accepted Answer

The slope of the given line $18x + 9y = -37$ can be found by rewriting it in slope-intercept form $y = mx + b$. The equation becomes $y = -2x - \frac{37}{9}$, so the slope $m = -2$. A line perpendicular to this line will have a slope that is the negative reciprocal, which is $\frac{1}{2}$.To find the tangent line to the curve $y = \frac{1}{6}x^3 - \frac{1}{4}x^2 - x + 4$ that is perpendicular to the given line, we first find the derivative of the curve, which represents the slope of the tangent line at any point $x$. The derivative is $y' = \frac{1}{2}x^2 - \frac{1}{2}x - 1$.Setting this equal to the slope of the line perpendicular to the given line ($\frac{1}{2}$), we get $\frac{1}{2}x^2 - \frac{1}{2}x - 1 = \frac{1}{2}$. Solving this quadratic equation for $x$ gives $x = 2$ and $x = -1$.Substituting these $x$ values back into the original equation of the curve to find the corresponding $y$ values, we get the points $\left( 2 , \frac{7}{3} \right)$ and $\left( -1 , \frac{55}{12} \right)$.

The Tangent Line to the Curve Y = 16\frac { 1 } { 6 }61​

The Tangent Line to the Curve Y = $\frac { 1 } { 6 }$