What is the equation of the line normal to #f(x)=2x^2-x - 1# at #x=5#?

1 Answer
Dec 25, 2015

#y = -1/19x +44 5/19 #

or

#y = -1/19 x + 841/19

Explanation:

Remember #m = f'(x)# is slope of tangent line

Slope of normal line is #m_2 = 1/m# as it is perpendicular to the tangent line

#f(x) = 2x^2 - x - 1# at #x= 5#

Step 1: Find derivative, to determine the slope of tangent line

#f'(x) = d/dx(2x^2) -d/dx(x) -d/dx(1)#

#f'(x) = 4x -1#

Step 2: Find the slope of the tangent line

#m= f'(5) = 4(5) -1 = 19#

Slope of normal line is #m_2 = (-1)/19#

Step 3 Determine the #y# coordinate, of #f(x)# , when #x= 5#

#f(5) = 2(5)^2 -5 -1#

#f(5) = 50-5-1#

#f(5) = 44#

Ordered pair #(5, 44)#

Step 4: Write the equation of the line using point slope formula

#m_2 = -1/19 , " " " (5, 44)#

#y -44 = -1/19(x-5) #

#y-44 = -1/19 x +5/19 #

#y = -1/19x + 5/19 +44#

#y = -1/19x +44 5/19 #

or
#y = -1/19 x + 841/19#