Question

suppose that $f(x)=2x^{2}+5$. (a) find the slope of the line tangent to $f(x)$ at $x = - 7$. (b) find the instantaneous rate of change of $f(x)$ at $x=-7$. (c) find the equation of the line tangent to $f(x)$ at $x = - 7. y=$

Explanation:

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

Using the power - rule, if $f(x)=2x^{2}+5$, then $f^\prime(x)=\frac{d}{dx}(2x^{2}+5)=4x$.

Step2: Find the slope of the tangent line at $x = - 7$

Substitute $x=-7$ into $f^\prime(x)$. So $m = f^\prime(-7)=4\times(-7)=-28$.

Step3: Find the value of $f(-7)$

$f(-7)=2\times(-7)^{2}+5=2\times49 + 5=98 + 5=103$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-7,103)$ and $m=-28$.
$y - 103=-28(x + 7)$
$y-103=-28x-196$
$y=-28x - 93$.

Answer:

(A) - 28
(B) - 28
(C) $y=-28x - 93$