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

Question

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

Emma Aldridge · Accepted Answer

#y = -1/3x+2/3# # f(x)=x^2-x-2 # at # x=2# we get #y=f(2) = 0# #f'(x) = 2x-1#, so the slope of the tangent at #(2,0)# is #f'(2)=3#. Therefore the slope of the normal line is #-1/3#. And the equation of the line through #(2,0)# with slope #-1/3# is #y = -1/3x+2/3#.

Emma Aldridge · Answer

undefined The equation of the line normal to f(x)=x^2-x-2 at x=2 is y = -3x + 8.

Emma Aldridge · Answer

undefined To find the equation of the line normal to $f(x) = x^2 - x - 2$ at $x = 2$, first, find the derivative of $f(x)$. Then, find the slope of the tangent line at $x = 2$. Since the line normal to the curve is perpendicular to the tangent line, its slope is the negative reciprocal of the slope of the tangent line. Finally, use the point-slope form to find the equation of the line.

1. Find $f'(x)$: 
$$f'(x) = 2x - 1$$

2. Evaluate $f'(2)$ to find the slope of the tangent line at $x = 2$:
$$f'(2) = 2(2) - 1 = 3$$

3. The slope of the line normal to the curve at $x = 2$ is the negative reciprocal of the slope of the tangent line:
$$m_{\text{normal}} = -\frac{1}{3}$$

4. Now, use the point-slope form with the point $(2, f(2))$ and the slope $m_{\text{normal}}$:
$$y - f(2) = m_{\text{normal}}(x - 2)$$

5. Evaluate $f(2)$:
$$f(2) = (2)^2 - 2(2) - 2 = 2$$

6. Substitute the values into the point-slope form:
$$y - 2 = -\frac{1}{3}(x - 2)$$

7. Simplify the equation to find the final answer:
$$y - 2 = -\frac{1}{3}x + \frac{2}{3}$$
$$y = -\frac{1}{3}x + \frac{8}{3}$$