# What is this question asking?

1. Apr 16, 2016

### Alexandra Fabiello

1. The problem statement, all variables and given/known data

Find the equation of the curve for which y'' = 8 if the curve is tangent to the line y = 11x at (2, 22).

2. Relevant equations

?

3. The attempt at a solution

y' = 8x + c1

y = 4x2 + c1x + c2

What exactly is the question asking me to do, especially with the y = 11x bit? ∫ydx?

2. Apr 16, 2016

### LCKurtz

It wants your curve to be tangent at (2,22). That means it must pass through that point and have the same slope there as the line.

3. Apr 16, 2016

### Alexandra Fabiello

But doesn't tangent mean derivative in this case? My curve should have a slope of 11 at (2,22), fine; how does that apply to y'' and y' and y as I said? My curve would be y, right, because y'' = 8 is the double derivative of that? How to solve it with two different constants, then?

4. Apr 16, 2016

### LCKurtz

You have two constants with which you can make the curve agree with point and slope.

5. Apr 16, 2016

### Alexandra Fabiello

So y = 4x2 + c1x + c2 = 11? Without knowing the slope for y', how can we solve for this?

I mean, so far, I've got -4x2 - c1x + 11 = c2, but now what?

6. Apr 17, 2016

### LCKurtz

You have $y = 4x^2 + c_1x + c_2$ with two unknown constants. What do the constants have to be so that $y(2) = 22$ and $y'(2) = 11$?

7. Apr 17, 2016

### Alexandra Fabiello

Oh, y' = 11 as well? Because y = 11x is a line?

8. Apr 17, 2016

### LCKurtz

There's really nothing more to tell you without working it for you. Just do it.

9. Apr 17, 2016

### Alexandra Fabiello

Got it. WileyPlus accepted my answer.

So basically, being tangent to a line means y' = 11 at that x-value, so it's solvable. If someone else asks a similar question, answer like that, and it might just make a lot more sense to them.