Determine where, if any where, the tangent line to f(x) =x^3 - 5x^2 +x is parallel to the line y = 4x +23.

I derived it to be 3x^2 - 10x + 1 . But not sure what to do next

Question

Determine where, if any where, the tangent line to f(x) =x^3 - 5x^2 +x is parallel to the line y = 4x +23.

I derived it to be 3x^2 - 10x + 1 . But not sure what to do next

anonymous · Answer

well u should solve the equation $$3x^2 - 10x + 1=4$$why?

3psilon · Answer

Why not = to 4x + 23?

anonymous · Answer

actually derivative of a function in the given point equals to the slope of tangent line in that point

3psilon · Answer

Wait I don't quite understand

anonymous · Answer

what @mukushla means, is that the slope of the  y = 4x +23 is equal to 4. The derivative you took is the slope of the tangent line. Si since you looking for a paralel to y = 4x +23, this derivative should be equal to 4 . So solve  3x^2 - 10x + 1 =4  for x to find which x gives you this slope.

3psilon · Answer

@myko I got X1 = -2.7 and x2=3.6

3psilon · Answer

Am I on the right track @mukushla

anonymous · Answer

x1 is not right i think

anonymous · Answer

so..at the points (x1,y1),(x2,y2)........
the curve has two tangents parallel to the given .line.........
right??!
@mukushla

3psilon · Answer

But I use the quadratic and keep getting that answer

anonymous · Answer

yes anusha

anonymous · Answer

u have$$3x^2-10x-3=0$$it gives: $$x_1=-0.27 \ \ \ x_2=3.61$$

3psilon · Answer

Oh sorry mine was a typo. I meant -.27 But what do you do next?

3psilon · Answer

@mukushla

anonymous · Answer

That's it. You found two points where tangent line to the curve f(x) =x^3 - 5x^2 +x has same slope (is parallel) to y = 4x +23