What is the equation of the line normal to #f(x)=2x^2-x+5 # at #x=-2#?

Question

What is the equation of the line normal to #f(x)=2x^2-x+5   # at #x=-2#?

Charles Arocha · Accepted Answer

The equation of the line will be #y = 1/9x + 137/9#.

Tangent is when the derivative is zero. That is #4x - 1 = 0. x = 1/4# At x = -2, f' = -9, so the slope of the normal is 1/9. Since the line goes through #x=-2# its equation is #y = -1/9x + 2/9# First we need to know the value of the function at #x = -2# #f(-2) = 2*4 + 2 + 5 = 15# So our point of interest is #(-2, 15)#. Now we need to know the derivative of the function: #f'(x) = 4x - 1# And finally we'll need the value of the derivative at #x = -2#: #f'(-2) = -9# The number #-9# would be the slope of the line tangent (that is, parallel) to the curve at the point #(-2, 15)#. We need the line perpendicular (normal) to that line. A perpendicular line will a negative reciprocal slope. If #m_(||)# is the slope parallel to the function, then the slope normal to the function #m# will be: #m = - 1/(m_(||))# This means the slope of our line will be #1/9#. Knowing this we can proceed with solving for our line. We know it will be of the form #y = mx + b# and will pass through #(-2, 15)#, so: #15 = (1/9)(-2) + b# #15 + 2/9 = b# #(135/9) + 2/9 = b# #b = 137/9# This means our line has the equation: #y = 1/9x + 137/9#

Elijah Abram · Answer

undefined To find the equation of the line normal to f(x) at x=-2, we need to determine the slope of the normal line and a point on the line.

First, we find the derivative of f(x) to obtain the slope of the tangent line at x=-2.

f'(x) = 4x - 1

Substituting x=-2 into f'(x), we get:

f'(-2) = 4(-2) - 1 = -9

The slope of the tangent line at x=-2 is -9.

Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of -9, which is 1/9.

Next, we find the y-coordinate of the point on the normal line by substituting x=-2 into f(x):

f(-2) = 2(-2)^2 - (-2) + 5 = 17

Therefore, the point on the normal line is (-2, 17).

Using the point-slope form of a line, we can write the equation of the line normal to f(x) at x=-2:

y - 17 = (1/9)(x + 2)