a. find an equation of the tangent line at x = a\nb. use a graphing utility to graph the curve and the…

Answer:

$y = 3x + 1$

Explanation:

Step1: Find the derivative of the function

The derivative of $y = x^{3}-3x^{2}+4x + 2$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=3x^{2}-6x + 4$.

Step2: Evaluate the derivative at $x = a$

Substitute $x = 1$ into $y'$. So $y'(1)=3(1)^{2}-6(1)+4=3 - 6 + 4 = 1$. This is the slope $m$ of the tangent line.

Step3: Find the y - coordinate of the point on the curve at $x = a$

Substitute $x = 1$ into $y=x^{3}-3x^{2}+4x + 2$. Then $y(1)=(1)^{3}-3(1)^{2}+4(1)+2=1 - 3+4 + 2=4$. So the point $(x_0,y_0)$ on the curve is $(1,4)$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_0=m(x - x_0)$. Substitute $m = 1$, $x_0 = 1$ and $y_0 = 4$ into it: $y-4=1\times(x - 1)$. Simplify to get $y=x - 1+4$, which is $y = 3x + 1$.