What is the equation of the line normal to #f(x)=-2x^2 +3x - 4 # at #x=-2#?

Question

Evelyn Agard · Accepted Answer

#y=-1/11x-18 2/11#

Given -

#y=-2x^2+3x-4#

The first derivative gives the slope at any given point

#dy/dx=-4x+3#

At #x=-2# its slope is

#dy/dx=-4(-2)+3=8+3=11#

It can be taken as the slope of the Tangent, drawn to that point on the curve.

Y- co-ordinate of the point -

#y=-2(-2)^2+3(-2)-4#
#y=-8-6-4=-18#

The point on the curve, the normal passing through is #(-2, -18)#

If the two lines are perpendicular then -

#m_2 xxm_2=-1#
#11 xx m_2=-1#
#m_2=(-1)/11#

The equation of the normal is -

#mx+c = y#
#-1/11(-2)+c=-18#
#2/11+c=-18#
#c=-18-2/11=18 2/11#

Equation is -

#y=-1/11x-18 2/11#

Cameron Alexander · Answer

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

First, we find the derivative of f(x) with respect to x, which gives us f'(x).

f'(x) = -4x + 3

Next, we substitute x=-2 into f'(x) to find the slope at x=-2.

f'(-2) = -4(-2) + 3 = 11

The slope of the line normal to f(x) at x=-2 is 11.

Using the point-slope form of a linear equation, we can write the equation of the line as:

y - y1 = m(x - x1)

Substituting the values x1=-2, y1=f(-2)=-2(-2)^2 + 3(-2) - 4 = -10, and m=11, we have:

y - (-10) = 11(x - (-2))

Simplifying, we get:

y + 10 = 11(x + 2)

Expanding and rearranging, the equation of the line normal to f(x) at x=-2 is:

y = 11x + 12