# Thread: Finding Points on a Graph with Specific Tangent Lines

1. ## Finding Points on a Graph with Specific Tangent Lines

Determine points on the graph y= x^(3/2) - x^(1/2) where the tangent line is parallel to y=x+3.

This question is in the unit of my manual about derivatives, which gives me the idea that I have to somehow use this. I know the slope of y=x+3 is 1, which means the tangent lines of the points on that graph will have a slope of 1, but I don't know how to work backward from there. Any hints as to the process would be helpful. There are no relevant examples in my manual at all.

2. ## Re: Finding Points on a Graph with Specific Tangent Lines

Originally Posted by Pewter12
Determine points on the graph y= x^(3/2) - x^(1/2) where the tangent line is parallel to y=x+3.

This question is in the unit of my manual about derivatives, which gives me the idea that I have to somehow use this. I know the slope of y=x+3 is 1, which means the tangent lines of the points on that graph will have a slope of 1, but I don't know how to work backward from there. Any hints as to the process would be helpful. There are no relevant examples in my manual at all.
$y' = \frac{3\sqrt{x}}{2} - \frac{1}{2\sqrt{x}} = 1$

$\frac{3x}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} = \frac{2\sqrt{x}}{2\sqrt{x}}$

$3x - 1 = 2\sqrt{x}$

$9x^2 - 6x + 1 = 4x$

$9x^2 - 10x + 1 = 0$

solve the quadratic for possible values of x on the curve ... don't forget to check for extraneous solutions.

3. ## Re: Finding Points on a Graph with Specific Tangent Lines

Originally Posted by Pewter12
Determine points on the graph y= x^(3/2) - x^(1/2) where the tangent line is parallel to y=x+3.

This question is in the unit of my manual about derivatives, which gives me the idea that I have to somehow use this. I know the slope of y=x+3 is 1, which means the tangent lines of the points on that graph will have a slope of 1, but I don't know how to work backward from there. Any hints as to the process would be helpful. There are no relevant examples in my manual at all.
How is the derivative of a function related to the function?

4. ## Re: Finding Points on a Graph with Specific Tangent Lines

Thank you for the replies. Is there a specific rule that allows me to ignore the denominator in the 2nd-3rd step? I would have thought I needed to multiply all the numerators by the denominator in order to go on to the next step. This is probably something simple that I'm missing, but would someone enlighten me?

5. ## Re: Finding Points on a Graph with Specific Tangent Lines

Originally Posted by Pewter12
Thank you for the replies. Is there a specific rule that allows me to ignore the denominator in the 2nd-3rd step? I would have thought I needed to multiply all the numerators by the denominator in order to go on to the next step. This is probably something simple that I'm missing, but would someone enlighten me?
$\frac{3x}{2\sqrt{x}} - \frac{1}{2\sqrt{x}} = \frac{2\sqrt{x}}{2\sqrt{x}}$ $\equiv$ $\frac{3x-1}{2\sqrt{x}} = \frac{2\sqrt{x}}{2\sqrt{x}}$

so then you can just simply set the numerators equal to each other since you have a common denominator. Make sense?

6. ## Re: Finding Points on a Graph with Specific Tangent Lines

What Youkla is saying is that if $\frac{a}{c}= \frac{b}{c}$ then a= b. One way to see that is to multuply on both sides by c (c can't be 0 because it in the denominator).

7. ## Re: Finding Points on a Graph with Specific Tangent Lines

Alright, thank-you, I suppose that makes sense. Now, I am just wondering for the step where the entire equation is squared, is there also a certain rule that allows me to manipulate the equation in that way?

8. ## Re: Finding Points on a Graph with Specific Tangent Lines

Originally Posted by Pewter12
Alright, thank-you, I suppose that makes sense. Now, I am just wondering for the step where the entire equation is squared, is there also a certain rule that allows me to manipulate the equation in that way?
If you are referring to going from this:

$3x - 1 = 2\sqrt{x}$

to this:

$9x^2 - 6x + 1 = 4x$

you eliminate the square root symbol by squaring both sides. That is a standard approach when solving for equations with a square root symbol in it.

9. ## Re: Finding Points on a Graph with Specific Tangent Lines

Alright, thank-you. I guess I was just uneasy about accepting that with the denominator still in mind. However, I suppose that if the common denominator is irrelevant then this should not matter. Thank-you for all of your clarification.