Find the equation of the tangent line to the curve when x has the given value.
-$$f ( x ) = 5 x ^ { 2 } + x ; x = - 4$$
A) $y = 13 x - 16$
B) $y = - \frac { 4 x } { 25 } + \frac { 8 } { 5 }$
C) $y = \frac { x } { 20 } + \frac { 1 } { 5 }$
D) $y = - 39 x - 80$

Question

Quizplus · Accepted Answer

To find the equation of the tangent line to the curve at a given point, we first need to find the derivative of the function, which gives us the slope of the tangent line at any point $x$. For $f(x) = 5x^2 + x$, the derivative $f'(x) = 10x + 1$. Evaluating this derivative at $x = -4$ gives us the slope of the tangent line at that point: $f'(-4) = 10(-4) + 1 = -39$. Next, we find the y-coordinate of the point on the curve where $x = -4$ by substituting $x = -4$ into the original function: $f(-4) = 5(-4)^2 + (-4) = 80 - 4 = 76$.Now, we have a point $(-4, 76)$ and a slope $-39$. 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 on the line. Substituting the values, we get $y - 76 = -39(x + 4)$. Simplifying, we find $y = -39x - 80$, which matches option D.

Find the Equation of the Tangent Line to the Curve f(x)=5x2+x;x=−4f ( x ) = 5 x ^ { 2 } + x ; x = - 4f(x)=5x2+x;x=−4

Find the Equation of the Tangent Line to the Curve $f ( x ) = 5 x ^ { 2 } + x ; x = - 4$