What is the equation of the normal line of f(x)= (x^2 + 2x - 1)/(4 - 3x) at x=-1?

1 Answer
Apr 5, 2017

y-49/6x-331/42=0 or

42y-343x-331=0

Explanation:

Before beginning, you should know that derivative of a function of

the form g(x)/f(x) is given by (f(x)*g'(x)-g(x)*f'(x))/(f(x))^2

i.e. if a(x)=g(x)/f(x) then,

color(red)(d/dxa(x) = (f(x)*d/dxg(x)-g(x)*d/dxf(x))/(f(x))^2)

Coming back to the question,

At x=-1; f(x)=(1-2-1)/(4+3)=-2/7

therefore we have to find equation of normal at (-1, -2/7)

slope m_1 of f(x) at any general point (x, y) is given by f'(x)

i.e. m_1 = f'(x)

= ((4-3x)*d/dx(x^2+2x-1)-(x^2+2x-1)*d/dx(4-3x))/(4-3x)^2

=((4-3x)(2x+2)-(x^2+2x-1)(-3))/(4-3x)^2

therefore slope of f(x) at x=-1 is given by m_1]_(x=-1)

m_1]_(x=-1)=((4+3)(-cancel(2)+cancel(2))-(1-2-1)(-3))/(4+3)^2

implies m_1]_(x=-1)= (3*(-2))/49 = -6/49

Let the slope of normal at the given point be m_2.

When two curves are normal/perpendicular to each other at some point then the product of their slopes at that point equals -1.

therefore at (-1, -2/7); m_1*m_2=-1 implies m_2=-1/m_1 = 49/6

Hence equation of normal at given point can be written in slope point form as

y-(-2/7)=m_2(x-(-1))

y+2/7=49/6*(x+1) = 49/6x+49/6

y-49/6x-331/42=0 is the required equation of normal.