Question

question
find the equation of the tangent line to the function f(x) = 2x^3 + 4x^2 - x + 2 at the point (-1, 5). give your answer in the form y = mx + b.
provide your answer below

Explanation:

Step1: Find the derivative of the function

Using the power - rule $(x^n)'=nx^{n - 1}$, if $f(x)=2x^{3}+4x^{2}-x + 2$, then $f'(x)=6x^{2}+8x-1$.

Step2: Evaluate the derivative at the given x - value

Substitute $x=-1$ into $f'(x)$. So $f'(-1)=6(-1)^{2}+8(-1)-1=6 - 8 - 1=-3$. The slope $m$ of the tangent line is $-3$.

Step3: Use the point - slope form to find the equation of the line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-1,5)$ and $m=-3$. So $y - 5=-3(x+1)$.

Step4: Rewrite the equation in slope - intercept form

Expand the right - hand side: $y - 5=-3x-3$. Then add 5 to both sides to get $y=-3x + 2$.

Answer:

$y=-3x + 2$