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

##### 1 Answer

May 27, 2016

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

#### Explanation:

Given -

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

The first derivative gives the slope at any given point

#dy/dx=-4x+3#

At

#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

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#