# Thread: How to find equation of tangent of function

1. ## How to find equation of tangent of function

Given g(x) = √(X-1), find the equation of tangent line to the graph at the point (5,2).

I have no idea why this was given as an assignment; this is not something I have any recollection of learning how to do.

Thank you.

2. let $y = mx + n$ be the equation of the tangent line we are looking for.

First, you know the line passes through the point (5,2) -- so you can put x=5, y=2 and get your first equation.

Second, you know the line $y=mx+n$ is tangent to the curve $g(x) = \sqrt{x-1}$ at the point (5,2) -- that means that they have the same slope at x=5, or more precisely that $g'(5) = m$.

Can you follow with the rest?

I'm not sure I'm following, should I plug in 5 in my function?

That would give me g(5) = 2? So that my slope would be 2.

That would give me y= 2X - 8 which doesn't really make sense.

There must be something I don't get.

4. Originally Posted by Mrs. White

I'm not sure I'm following, should I plug in 5 in my function?

That would give me g(5) = 2? So that my slope would be 2.

That would give me y= 2X - 8 which doesn't really make sense.

There must be something I don't get.
g(5) = 2 is the point on the curve which the tangent line "touches".

the slope of that line is g'(5) ... know what that means?

5. Originally Posted by skeeter
the slope of that line is g'(5) ... know what that means?
No I don't. How am I to find the equation of this line if the slope is g'(5)?

Thank you so much.

6. Originally Posted by Mrs. White
No I don't. How am I to find the equation of this line if the slope is g'(5)?

Thank you so much.
$g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x-5}$

7. Originally Posted by skeeter
$g'(5) = \lim_{x \to 5} \frac{g(x) - g(5)}{x-5}$

Does that mean I would replace h by 5 in the difference quotient equation I found?

This is what I found:
1/√(x+h)-1 + √(x-1)

So now I would do

√(5+5)-1 + √(5-1) = √9 + √4 = 5, which would mean my slope is 1/5 ?

If so my equation would be: Y= 1/5(X) +1 ??

I hope this is correct, that would be such a relief.

8. Originally Posted by Mrs. White
Does that mean I would replace h by 5 in the difference quotient equation I found?

This is what I found:
1/√(x+h)-1 + √(x-1)

So now I would do

√(5+5)-1 + √(5-1) = √9 + √4 = 5, which would mean my slope is 1/5 ?

If so my equation would be: Y= 1/5(X) +1 ??

I hope this is correct, that would be such a relief.

sorry ... but that is not how a limit of a difference quotient is determined.

if you do it correctly, you will find that the the slope of $g(x)$ at $x = 5$ is $g'(5) = \frac{1}{4}$

9. Originally Posted by skeeter
sorry ... but that is not how a limit of a difference quotient is determined.

if you do it correctly, you will find that the the slope of $g(x)$ at $x = 5$ is $g'(5) = \frac{1}{4}$
This is what I had previously found ! But I thought I was wrong. I plugged in 0 for h and it gave me a slope of 1/4.

My equation was then y=1/4X + 3/4.

Was that the correct way to proceed (plug in 5 for x and 0 for h)?