Another Tangent Line Question

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Scrutinize

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I understand completely what the question is asking me to do, I'm just unable to figure out the best way to solve it. Any ideas would be extremely helpful, thanks!
 
One way would be to write the equation of the tangent to the curve at any point x=a, and then set this line to pass through the given point. Solve the resulting equation for a.
 
Ok, so I got that the derivative is (-32x -8)/((4x^2 - 4)^(3/2)) Now I know that that is a curve of all the instantaneous slopes on the original graph. But I'm confused on how to use this info to get the equation to solve the question. For example if you use point slope form and plug in the equation of the derivatives in for the slope, and use the point given, you still need another point because there is an x and a y as well as an x1 and an x2. You need 2 points. I'm just kind of confused on what to do, but I could be looking at it completely wrong still.
 
Scrutinize said:
Ok, so I got that the derivative is (-32x -8)/((4x^2 - 4)^(3/2)) Now I know that that is a curve of all the instantaneous slopes on the original graph. But I'm confused on how to use this info to get the equation to solve the question. For example if you use point slope form and plug in the equation of the derivatives in for the slope, and use the point given, you still need another point because there is an x and a y as well as an x1 and an x2. You need 2 points. I'm just kind of confused on what to do, but I could be looking at it completely wrong still.
Click to expand...
The slope of the tangent line (not the curve) is (-32x -8)/((4x^2 - 4)^(3/2)).
 
Let's begin by simplifying the given function:

[MATH]f(x)=\frac{x+4}{\sqrt{x^2-1}}[/MATH]
Following Dr. Peterson's suggestion, we may write:

[MATH]y=-\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(x-a)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Now, if this line is to pass through the given point, we have:

[MATH]0=\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(a+1)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Solve this for \(a\), keeping in mind the domain...what do you get?
 
Here is how I would think about doing this. Pick an arbitrary point (a, f(a)) on the curve. Then find the slope of the line from (a, f(a)) to (-1,0). I would then find the derivative of f(x) at x=a, that is find f' (a). Finally set f'(a) equal to the slope from (a, f(a)) to (-1,0) and solve for a.
 
MarkFL said:
Let's begin by simplifying the given function:

[MATH]f(x)=\frac{x+4}{\sqrt{x^2-1}}[/MATH]
Following Dr. Peterson's suggestion, we may write:

[MATH]y=-\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(x-a)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Now, if this line is to pass through the given point, we have:

[MATH]0=\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(a+1)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Solve this for \(a\), keeping in mind the domain...what do you get?
Click to expand...
From above (response #6)you can write:

(4a+1)*(a+1) + (a+4) *(a^2-1) = 0

Simplify and solve for 'a'
 
So I just came back to this problem and I can’t find my work and I completely forgot how I did it and I don’t really understand what you guys said up there could someone explain it to me as if I was a newborn baby please. Thank you so much in advance and sorry for not understanding.
 
A newborn baby? Do you have about 18 years for us to explain?

How about showing what you tried this time, so we can have a head start?
 
Dr.Peterson said:
A newborn baby? Do you have about 18 years for us to explain?

How about showing what you tried this time, so we can have a head start?
Click to expand...
I meant, like explain it in like simpler terms. I'm getting lost in your explanation. I'm at the same point that I messaged about up there. I have the derivative and I know what its asking I just can't figure out how i'm supposed to solve it.
 
Scrutinize said:
I meant, like explain it in like simpler terms. I'm getting lost in your explanation. I'm at the same point that I messaged about up there. I have the derivative and I know what its asking I just can't figure out how i'm supposed to solve it.
Click to expand...
Please read carefully response #2 (edited) & #6 and tell us EXACTLY where you are getting lost!
 
Last edited by a moderator:
Dr.Peterson said:
One way would be to write the equation of the tangent to the curve at any point x=a, and then set this line to pass through the given point. Solve the resulting equation for a.
Click to expand...


So I'm confused as to what I'm replacing with a. If I'm just replacing all the x's with a's then what am I doing? And then I'm confused on what you're setting it equal to?

Then


MarkFL said:
Let's begin by simplifying the given function:

[MATH]f(x)=\frac{x+4}{\sqrt{x^2-1}}[/MATH]
Following Dr. Peterson's suggestion, we may write:

[MATH]y=-\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(x-a)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Now, if this line is to pass through the given point, we have:

[MATH]0=\frac{4a+1}{(a^2-1)^{\frac{3}{2}}}(a+1)+\frac{a+4}{\sqrt{a^2-1}}[/MATH]
Solve this for \(a\), keeping in mind the domain...what do you get?
Click to expand...

I see that that is the point slope form, but what is being plugged into the equation. Like what was replaced with a and what wasn't and why did we only replace some of them?
 
Scrutinize said:
So I'm confused as to what I'm replacing with a. If I'm just replacing all the x's with a's then what am I doing? And then I'm confused on what you're setting it equal to?

Then




I see that that is the point slope form, but what is being plugged into the equation. Like what was replaced with a and what wasn't and why did we only replace some of them?
Click to expand...
You did not read response #2 carefully. It states:

One way would be to write the equation of the tangent to the curve at any point x=a, and then set this line to pass through the given point. Solve the resulting equation for a.​
​
Use pencil-paper and write all the steps down - don't just stare at the screen!
 
The problem is:

Given the function: [MATH]f(x) = \frac{2x+8}{\sqrt{4x^2-4}}[/MATH].​
Find the slope of the tangent line to the function that passes through the point (-1,0) that is not on the graph of the function.​

As MarkFL pointed out, the function simplifies to [MATH]f(x) = \frac{x+4}{\sqrt{x^2-1}}[/MATH].

You have found that the derivative is [MATH]f'(x) = \frac{4x+1}{(x^2-1)^{3/2}}[/MATH].

Now we want to find a point [MATH](a, f(a))[/MATH] on the curve such that the tangent line passes through the given point [MATH](-1, 0)[/MATH]. [The problem just points out that this point is not on the graph of the function, which really doesn't matter.]

The slope of the tangent line at [MATH]x=a[/MATH] is [MATH]f'(a)[/MATH]. That is the "m" you want to put into the point-slope form. So, yes, you do want to replace [MATH]x[/MATH] in the derivative with [MATH]a[/MATH]. Then you put the resulting expression for the slope in place of the m in [MATH]y - y_1 = m(x - x_1)[/MATH]. That gives the equation of the tangent line.

Then you want [MATH](-1, 0)[/MATH] to be on this line; so replace x and y in the equation of the tangent line with -1 and 0, respectively.

The result is an equation with one variable, namely a. (It's the equation MarkFL showed you.

Solve this for a. Then you can put that back into the formula for the slope, which will be the answer to the question.

There are other ways to do this (some perhaps a littler cleaner), but we're sticking with this one to avoid complicating things.
 
So I understand what to substitute in for x and y but what am I putting in for x1 and y1, namely for y1 specifically. I understand that x1 would just be a but what do i plug in for y1?
 
Also I got a^3 - 6a -5 = 0 but when I solve it I don't get what mark got up there? I get 1+ 21^(1/2) divided by two and the same but minus which gives me two different answers. WHen I plugged in the equation I got into the calculator I got the same answer as that up there. I copied the equation given up there at response #6 so im really confused now.
 
What do [MATH]x_1[/MATH] and [MATH]y_1[/MATH] mean in the formula? They are the starting point for the line, which (as I was presenting it) is the point [MATH](a, f(a))[/MATH]. That is, [MATH]y_1[/MATH] is the value of [MATH]y[/MATH] on the graph when [MATH]x = a[/MATH].

But if that's all too confusing, try the method I actually used when I solved the problem: You have two points that have to be on the line, namely [MATH](a, f(a))[/MATH], which specifically is [MATH]\left(a, \frac{a+4}{\sqrt{a^2-1}}\right)[/MATH], and [MATH](-1, 0)[/MATH]. You know the slope of the line, [MATH]m=\frac{4a+1}{(a^2-1)^{3/2}}[/MATH].

Now put those two points and that slope into the formula for the slope between two points, [MATH]m = \frac{y_2 - y_1}{x_2 - x_1}[/MATH]. Then solve that equation.
 
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