Can you find the equation to the curve and show steps. See below

A function f has 2nd deriavative f"(x)=6(x-1). Find the function if its graph passes through the pt (2,1) & at that point is tangent to the line given by 3x-y-5=0.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

We are given that `f''(x)=6(x-1)` , that the function has the point (2,1), and that the function is tangent to the line 3x-y-5=0 at (2,1).

So `f''(x)=6x-6`

Then `f'(x)=3x^2-6x+k` by taking an antiderivative.

Now the function is tangent to the line y=3x-5, so the slope of the tangent line is 3 at the point (2,1). Thus `f'(2)=3` .


Now `f'(x)=3x^2-6x+3`

So `f(x)=x^3-3x^2+3x+k`

Since (2,1) lies on the curve, f(2)=1.

`(2)^3-3(2)^2+3(2)+k=1 ==>k=-1`


The equation of the curve is `f(x)=x^3-3x^2+3x-1`


The graph of the curve and the tangent line (in red):

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial