Question

suppose that (f(x)=5x^{2}+3). (a) find the slope of the line tangent to (f(x)) at (x = - 2). (b) find the instantaneous rate of change of (f(x)) at (x=-2). (c) find the equation of the line tangent to (f(x)) at (x = - 2) (y=). note: you can earn partial credit on this problem.

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule, if $f(x)=5x^{2}+3$, then $f^\prime(x)=\frac{d}{dx}(5x^{2}+3)=10x$.

Step2: Find the slope of the tangent line at $x = - 2$ (Answer for part A and B)

Substitute $x=-2$ into $f^\prime(x)$. So $m = f^\prime(-2)=10\times(-2)=-20$.

Step3: Find the value of $y$ at $x=-2$ for $f(x)$

Substitute $x = - 2$ into $f(x)$. $y=f(-2)=5\times(-2)^{2}+3=5\times4 + 3=20 + 3=23$.

Step4: Find the equation of the tangent line (Answer for part C)

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-2,23)$ and $m=-20$.
$y - 23=-20(x + 2)$
$y-23=-20x-40$
$y=-20x - 17$.

Answer:

(A) - 20
(B) - 20
(C) $y=-20x - 17$