# Thread: Tangent to Given Curve

1. ## Tangent to Given Curve

Need help on this problem

have 0 idea what to do/how to start it...

2. Originally Posted by stiitches
Need help on this problem

have 0 idea what to do/how to start it...
tangent to the given curve at what point?

anyway, remember, the derivative gives the slope of the tangent line at any point.

3. Originally Posted by Jhevon
tangent to the given curve at what point?

anyway, remember, the derivative gives the slope of the tangent line at any point.
I dont know, thats the whole question...
So I should start by finding the derivative of the first function? or the second one?

4. Originally Posted by stiitches
I dont know, thats the whole question...
So I should start by finding the derivative of the first function? or the second one?
In order for a line to be tangent to a curve, at a given point, both curve and line must pass through the point and must have the same derivative there.

So you need to find c and x such that $y= \frac{5}{6}x+ \frac{15}{2}= c\sqrt{x}$ and they have the same derivative for that x. That is two equations to solve for x and c.

I will give you one derivative free- the derivative of $y= \frac{5}{6}x+ \frac{15}{2}$ is $\frac{5}{6}$, for all x. What is the derivative of $c\sqrt{x}$? (That will depend on x.)