# Homework Help: Finding the point on the graph where the tangent is parallel

1. Dec 16, 2008

### meeklobraca

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

Find the point on the graph y = sqrt 2x-1 where the tangent line is parallel to x - 3y = 16

2. Relevant equations

3. The attempt at a solution

So what I started off was was trying to find the slope of x - 3y = 16. I rearranged the equation to a y = equation and got 16-x /-3 which I got a slope of -3.

Then I went to find the deriviative of y = sqrt 2x-1 which I get (x - 1/2)exp-1/2

From there i get a little lost. The derivative = the slope of the tangent which would get me my x and y points, but my x values get a lot wonky. Would somebody be able to lead me in the right direction of finding my x point?

Thanks!

2. Dec 16, 2008

### gabbagabbahey

Be careful with your negative signs!

$$y=16-\frac{x}{-3}=16+\frac{x}{3}$$

You should double check your derivative too!

3. Dec 16, 2008

### meeklobraca

OKay, with that then I get a slope of 1/3, im still not clear on the derivative though? If its
(2x-1)exp 1/2, wouldnt that be (1/2)(2x-1)exp-1/2?

4. Dec 16, 2008

### gabbagabbahey

You need to use the chain rule:

$$\frac{d}{dx} (2x-1)^{1/2}=\frac{1}{2}(2x-1)^{-1/2}\cdot\frac{d}{dx}(2x-1)$$

5. Dec 16, 2008

### meeklobraca

Okay, this is what ive got from your equation.

1/2(2x-1)-1/2 x 2

(2x-1)-1/2

So with that then I have the deriviative equaling the slope of the tangent which ive figured to be 1/3.

How do I get rid of that negative exponent?

6. Dec 16, 2008

### meeklobraca

To answer my own question, I took the reciprocal of the 2x-1 line so I got

1/(2x-1) = 1/9 A little cross multiplication and I landed x = 5?

Throw that into the equation and we get a y value of 3, therefore the answer is (5,3)?

haha, sorry bud im editting this as your trying to reply, it dawned on me in like 5 seconds, what did I do with the square root??

Last edited: Dec 16, 2008
7. Dec 16, 2008

### gabbagabbahey

Looks good to me

8. Dec 16, 2008

### meeklobraca

9. Dec 16, 2008

### HallsofIvy

He was assuming you meant what you said!

16- x/-3 means 16- (x/-3). If you mean (16- x)/-3 you should have used parentheses to make that clear.

10. Dec 16, 2008

### meeklobraca

So which is it that I go from?

If you have x - 3y = 16 how do you adjust that to make it a y = equation?

11. Dec 16, 2008

### Dick

x-3y=16 -> x-16=3y -> (x-16)/3=y -> x/3-16/3=y. This shouldn't be the hardest part of the problem. gabba didn't check the derivation from the start.

12. Dec 17, 2008

### meeklobraca

If you wouldnt mind confirming for me, but I dont think that changes anything, my answer of (5,10) is still correct?

13. Dec 17, 2008

### Dick

x=5 is good. But if y=sqrt(2x-1) what does that make y?

14. Dec 17, 2008