What is the equation of the line that is normal to #f(x)= (x-3)^2-2x-2 # at # x=-1 #?

Question

What is the equation of the line that is normal to #f(x)=  (x-3)^2-2x-2 # at #  x=-1  #?

Emma Aldridge · Accepted Answer

# 6x-y+22=0# Lets differentiate the function to find the slope of the tangent at #x=-1#, then we can find slope of the normal and find the equation of it. Differentiating the function w.r.t #x#, we get, #f'(x)=2(x-3)-2# We find the slope of the tangent at # x=-1#, we get #f'(-1)=-6# Slope of the tangent is -6, hence slope of the normal is 6 , this because the product of slope of tangent and slope of normal at a point is #-1#. Before finding the equation, we need the value of #f(x)# at #x=-1#, #f(-1)=16# Now we have the point #(-1,16)# and slope = #6#, which enough data to find the equation of the normal. The equation of the normal is #(y-16)=6(x+1)# On simplification, we get # 6x-y+22=0#

Ezekiel Alexander · Answer

undefined To find the equation of the line that is normal to the function f(x) = (x-3)^2 - 2x - 2 at x = -1, we need to determine the slope of the tangent line at x = -1 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of the function f(x) with respect to x and evaluate it at x = -1.

The derivative of f(x) = (x-3)^2 - 2x - 2 is f'(x) = 2(x-3) - 2.

Evaluating f'(x) at x = -1, we get f'(-1) = 2(-1-3) - 2 = -12.

The slope of the tangent line at x = -1 is -12.

To find the slope of the normal line, we take the negative reciprocal of -12, which is 1/12.

Now, we have the slope of the normal line, and we also know that it passes through the point (-1, f(-1)).

Substituting x = -1 into the original function f(x), we get f(-1) = (-1-3)^2 - 2(-1) - 2 = 16 + 2 - 2 = 16.

Therefore, the point of intersection is (-1, 16).

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of intersection and m is the slope of the normal line, we can substitute the values to find the equation of the line.

Plugging in the values, we get y - 16 = (1/12)(x - (-1)).

Simplifying, we have y - 16 = (1/12)(x + 1).

Expanding, we get y - 16 = (1/12)x + 1/12.

Finally, rearranging the equation, we have y = (1/12)x + 1/12 + 16.

Simplifying further, the equation of the line that is normal to f(x) = (x-3)^2 - 2x - 2 at x = -1 is y = (1/12)x + 193/12.