What is the equation of the line normal to # f(x)=(x-4)(x-2)+2x^2-3x+4# at # x=0#?

1 Answer
Dec 3, 2015

#y = 1/9x + 12#

Explanation:

Let's simplify the function first.

#f(x) = (x-4)(x-2) + 2x^2 - 3x + 4#

#color(white)(xxx) = x^2 - 6x + 8 + 2x^2 - 3x + 4#

#color(white)(xxx) = 3 x^2 - 9x + 12#

As next, to find the slope of the tangent and normal line, we need to compute the derivative of the function:

#f'(x) = 6x - 9#

Evaluating the derivative at #x=0# gives you the slope of the tangent at #x = 0#:

#m_t = f'(0) = -9#

The slope of the normal line at #x = 0# can be computed as

#m_n = - 1 / m_t = 1/9#

Now, we have the slope of the normal line and we have the #x# value of a point on this line: #x = 0#. Let's find the according #y# value by evaluating #f(0)#:

#f(0) = 3 * 0 - 9 * 0 + 12 = 12#

The normal line can be described via the line equation

#y = m_n * x + n#

Plug the values #m_n = 1/9#, #x = 0# and #y = 12# to find #n#:

#12 = 1/9 * 0 + n#

#=> n = 12#

So, the equation of the normal line is

#y = 1/9x + 12#