Question

f(x)=3x^{2}-x + 4 at x=-1
enter the equation of the tangent line here.
y=-7x + 1
enter the equation of the normal line here.
y + 8=\frac{1}{7}(x + 1)

Explanation:

Step1: Find the derivative of the function

The derivative of $f(x)=3x^{2}-x + 4$ is $f^\prime(x)=6x-1$.

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

Substitute $x=-1$ into $f^\prime(x)$: $f^\prime(-1)=6\times(-1)-1=-6 - 1=-7$. This is the slope of the tangent line.

Step3: Find the point on the function at $x=-1$

Substitute $x = - 1$ into $f(x)$: $f(-1)=3\times(-1)^{2}-(-1)+4=3 + 1+4=8$. So the point is $(-1,8)$.

Step4: Find the equation of the normal line

The slope of the normal line is the negative - reciprocal of the slope of the tangent line. Since the slope of the tangent line is $-7$, the slope of the normal line is $\frac{1}{7}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(-1,8)$ and $m=\frac{1}{7}$, we get $y - 8=\frac{1}{7}(x + 1)$.

Answer:

The equation of the normal line should be $y-8=\frac{1}{7}(x + 1)$ instead of $y + 8=\frac{1}{7}(x + 1)$. The correct equation of the normal line is $y=\frac{1}{7}x+\frac{1}{7}+8=\frac{1}{7}x+\frac{57}{7}$, or in point - slope form $y - 8=\frac{1}{7}(x + 1)$.