What is the equation of the line normal to f(x)=x2−x at x=-2?

Question

Kaitlynn Craig · Accepted Answer

The normal line will intersect the curve at (-2, f(-2)). Since

f (- 2) = {(- 2)}^{2} - (- 2) = 6

, we know the normal line passes through the point (2, 6).
To find the slope of the normal line, first find the slope of the tangent line at that point by finding the value of the derivative at x=-2. Then, since the normal line is perpendicular to the tangent line, take the opposite reciprocal of the slope of the tangent line.
Through the power rule, we see that f'(x)=2x-1. We then see that the slope of the tangent line at x=-2 is f'(-2)=2(-2)-1=-5
The slope of the normal line is then

- \frac{1}{- 5} = \frac{1}{5}

Using the point (-2, 6) and slope of

\frac{1}{5}

, we can write the equation of the line from
y=mx+b

6 = \frac{1}{5} (- 2) + b \Rightarrow b = \frac{32}{5}

The normal line is:

y = \frac{1}{5} x + \frac{32}{5}

Vegljamzt6 · Accepted Answer

Solution: f(x)=x2−x and x0=−2.Find the value of the function at the given point: y0=f(−2)=6The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function, evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(x2−x)′=2x−1Hence, M(x0)=−1f′(x0)=−12x0−1Next, find the slope at the given point.m=M(−2)=15Finally, the equation of the normal line is y−y0=m(x−x0)Plugging the found values, we get that y−6=x−(−2)5Or, more simplify: y=x5+325.

What is the equation of the line normal

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