a. find the derivative function f’ for the function f. b. find an equation of the line tangent to the graph…

Answer:

b. $y = -x - 4$

Explanation:

Step1: Find $f(a)$

Substitute $a=-1$ into $f(x)=2x^{2}+3x - 2$. $f(-1)=2(-1)^{2}+3(-1)-2=2 - 3 - 2=-3$.

Step2: Find the slope of the tangent line

The slope of the tangent line at $x = a$ is $f^{\prime}(a)$. Since $f^{\prime}(x)=4x + 3$, when $a=-1$, $f^{\prime}(-1)=4(-1)+3=-1$.

Step3: Use the point - slope form

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(a,f(a))=(-1,-3)$ and $m = f^{\prime}(a)=-1$. $y-(-3)=-1(x - (-1))$.

Step4: Simplify the equation

$y + 3=-x - 1$, so $y=-x - 4$.