# Thread: Calc BC Curve of Tangent Line- Hor/Vert Problem

1. ## Calc BC Curve of Tangent Line- Hor/Vert Problem

My Calc teacher, well, he's cruel. Hard problem..

Given that $\displaystyle f(x) = 3x² + x / 2x -1$ , find all of the points on the curve where the tangent line has to be horizontal and vertical.

I first solved the derivative of the equation to get 6x ²-6x-1 / (2x-1)².

From there, do I solve where the numerator is 0 (horizontal) and where the denominator is 0 (vertical)?

Thank you. I seem stuck, as always...

2. Originally Posted by r2d2
My Calc teacher, well, he's cruel. Hard problem..

Given that $\displaystyle f(x) = 3x² + x / 2x -1$ , find all of the points on the curve where the tangent line has to be horizontal and vertical.

I first solved the derivative of the equation to get 6x ²-6x-1 / (2x-1)².

From there, do I solve where the numerator is 0 (horizontal) and where the denominator is 0 (vertical)?

Thank you. I seem stuck, as always...
Since you said he's cruel, he might be being extremely tricky because a tangent line cannot be both horizontal and vertical, but I'll assume he isn't quite that bad, anyway

you would do exactly what you thought... solve for when the denominator is 0 and the numerator is 0, but depending on how "cruel" he is, your answer for the denominator=0 will be x=1, which doesnt correspond to a point on the curve as it creates a vertical assymptote

3. Originally Posted by r2d2
My Calc teacher, well, he's cruel. Hard problem..

Given that $\displaystyle f(x) = 3x² + x / 2x -1$ , find all of the points on the curve where the tangent line has to be horizontal and vertical.

I first solved the derivative of the equation to get 6x ²-6x-1 / (2x-1)².

From there, do I solve where the numerator is 0 (horizontal) and where the denominator is 0 (vertical)?

Thank you. I seem stuck, as always...
So you want to find the horizontal tangents, which would be at the points where the derivative is zero. So yes, the numerator will be zero at those points. Now for the vertical tangents . I would think that would only occur at points of inflection. So find the second derviative and set that equal to zero.

So you want to find the horizontal tangents, which would be at the points where the derivative is zero. So yes, the numerator will be zero at those points. Now for the vertical tangents . I would think that would only occur at points of inflection. So find the second derviative and set that equal to zero.

No you are mistaking. Vertical tangents occur when the slope is $\displaystyle \infty$ or, i.e. when the denominator of the derivative =0

Though I suppose in this case you could argue about the point of inflection... which actually works for this particular case, but consider $\displaystyle \frac{1}{x^2}$ where there is no change in concavity yet vertical assymptotes

5. Originally Posted by artvandalay11
No you are mistaking. Vertical tangents occur when the slope is $\displaystyle \infty$ or, i.e. when the denominator of the derivative =0
Oh yess , sorry about that. I see why that would not be correct in general.

6. So I would set 6x²-6x-1 equal to 0 and solve for the x values? Any quick way to do that besides writing out the quad formula?

And when it says find the POINTS, would i plug those values for x into the original equation?

Oh, and would there be another way to get the vertical tangent lines besides taking the second derivative?
**Just saw the previous posts**How would I solve for the vertical line points?
Sorry, I am having trouble..

Thanks

7. Originally Posted by r2d2
So I would set 6x²-6x-1 equal to 0 and solve for the x values? Any quick way to do that besides writing out the quad formula?

And when it says find the POINTS, would i plug those values for x into the original equation?

Oh, and would there be another way to get the vertical tangent lines besides taking the second derivative?
**Just saw the previous posts**How would I solve for the vertical line points?
Sorry, I am having trouble..

Thanks

For horizontal tangents you need $\displaystyle 6x^2-6-1=0$ and you will need the quadratic formula

For vertical tangents, $\displaystyle (2x-1)^2=0$
then to find the points, plug in x and get the y values into f(x)

problem is, for the vertical tangents, $\displaystyle x=\frac{1}{2}$ is not actually a point on the curve

8. so to answer the vertical tangent question, I should only indicate that the x=1/2, because if i plug in 1/2 into the original equation , I get #/0, which does not exist?

9. since it says all the points on the curve where.... well there are no points on the curve with a vertical tangent, so i wouldn't consider it an answer