Find an equation of a line that is tangent to the graph of f and parallel to the given line.

Question

function: f(x)=x³ Line: 12x - y + 1 = 0

y (smaller y-intercept)=

y (larger y-intercept) =

I got -16 and 16, which was marked wrong. Please help. Thanks!

Answer 1

Hi Celine,

First step: As you know when two lines are parallel it means both have the same slope.

Second step: Given LIne is 12x - y + 1 = 0 ----> y = 12x +1 -----> slope is 12

Third step: For finding of slope of tangent line to the graph you need to derivative.

f(x) = x^3

f'(x)= 3x^2

3x^2 = slope = 12 ------> x^2 = 4 -----> x = ±2



if x = 2 -----> y = 2^3 =8 point is (2 , 8) equation of tangent line is y -8 = 12( x - 2) after simplify we have

y = 12x -16 and y-int = -16

when x = -2 -----> y = (--2)^3 = -8 point is (-2 , -8) and equation of tangent line is y -(-8) = 12 ( x -(-2)) after simplify we will have y = 12x +16 and y-int = 16

I hope it is useful,

Minoo