Expert: Sherry Wallin - 6/17/2009

Question
QUESTION: Find the equations for all the lines tangent to the graph of y = x3 – x that passes through the point (-2, 2).

ANSWER: Attir, need some help with calculus! To find the tangent line you need to take the first derivative and set it equal to zero and factor it to find the point of tangency:

Step 1:
Let f(x) = x^3-x so f'(x) = 3x^2-1 = 0-> 3x^2 = 1-> x^2 = 1/3->
x = +-sqrt(1/3)-> x = +-[sqrt(3)]/3

Step 2:
You need to find the line that goes through (-2,2) using the slopes in step 1 above:

The value of x above is your slopes and then use the point slope form of a line:

y-2= [(sqrt 3)/3](x+2)-> y = [(sqrt 3)/3]+(2/3)[(sqrt 3)+3]

and

y-2= [-(sqrt 3)/3](x+2)-> y = [-(sqrt 3)/3]-(2/3)[(sqrt 3)-3]

both lines which are tangent to y = x^3-x

Math Prof

---------- FOLLOW-UP ----------

QUESTION: i Sort out the answer like and the answer was:
y= -x
y=26x+54
is it ok.  

Answer
Attir~
    I'm not sure what it is is you are trying to ask? I gave you the equations of the lines that are tangent to y = x^3 – x that passes through the point (-2, 2).
They are: y = [(sqrt 3)/3]+(2/3)[(sqrt 3)+3]
and y = [-(sqrt 3)/3]-(2/3)[(sqrt 3)-3]

Both of these lines are tangent to y = x^3-x

Math Prof