Question

question: find the equation of the line tangent to the graph of f(x)=-x^2 + 4x + 2 at x=-1. sony, thats incorrect. try again? y = 7x feedback view answer submit switch current objective find the equation of a tangent line using basic derivative rules

Explanation:

Step1: Find the derivative of the function

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

Step2: Evaluate the derivative at $x=-1$

Substitute $x = - 1$ into $f'(x)$: $f'(-1)=-2\times(-1)+4=2 + 4=6$. The slope $m$ of the tangent line at $x=-1$ is 6.

Step3: Find the function value at $x=-1$

Substitute $x=-1$ into $f(x)$: $f(-1)=-(-1)^{2}+4\times(-1)+2=-1-4 + 2=-3$.

Step4: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-1,-3)$ and $m = 6$. So $y-(-3)=6(x-(-1))$, which simplifies to $y + 3=6(x + 1)$, and further to $y+3=6x+6$, and finally $y=6x + 3$.

Answer:

$y = 6x+3$