1. ## Limits

Find the equation of the tangent line to the curve $f(x) = \sqrt {x - 3}$ at the point $P(7,2)$.

$m = \lim_{x \to \-a} \frac{f(x) - f(a)}{x - a}$

$m = \lim_{x \to \-7} \frac{\sqrt{x - 3} - 2}{x - 7} \times \frac{\sqrt{x - 3} + 2}{\sqrt{x - 3} + 2}$

$m = \lim_{x \to \-7} \frac{x - 7}{(x - 7)(\sqrt {x - 3} + 2)}$

$m = \lim_{x \to \-7} \frac{1}{\sqrt {x - 3} + 2}$

$m = \frac {1}{4}$

$y - \frac {1}{4}x + b$

$2 = (\frac {1}{4})(7) + b$

$\frac {-3}{4} = b$

Therefore, $y = \frac {1}{4}x + \frac {-3}{4}$

$x - 4y + 1 = 0$

I'm confused? Don't know what I did wrong

2. ## I think

Originally Posted by Macleef
Find the equation of the tangent line to the curve $f(x) = \sqrt {x - 3}$ at the point $P(7,2)$.

$m = \lim_{x \to \-a} \frac{f(x) - f(a)}{x - a}$

$m = \lim_{x \to \-7} \frac{\sqrt{x - 3} - 2}{x - 7} \times \frac{\sqrt{x - 3} + 2}{\sqrt{x - 3} + 2}$

$m = \lim_{x \to \-7} \frac{x - 7}{(x - 7)(\sqrt {x - 3} + 2)}$

$m = \lim_{x \to \-7} \frac{1}{\sqrt {x - 3} + 2}$

$m = \frac {1}{4}$

$y - \frac {1}{4}x + b$

$2 = (\frac {1}{4})(7) + b$

$\frac {-3}{4} = b$

Therefore, $y = \frac {1}{4}x + \frac {-3}{4}$

$x - 4y + 1 = 0$

I'm confused? Don't know what I did wrong
That you confused the equation for a tangent line wrong $y-2=f'(7)(x-7)\Rightarrow{y-2=\frac{1}{4}\bigg(x-7\bigg)}$

3. ?? still confused, why does the equation contain y - 2 and x - 7? Isn't it just the slope equation that you use to find the equation of the tangent line to the curve?

4. ## Ok

Originally Posted by Macleef
?? still confused, why does the equation contain y - 2 and x - 7? Isn't it just the slope equation that you use to find the equation of the tangent line to the curve?
You I am sure have heard of the point slope form $y-y_1=m(x-x_1)$ yeah? so adapting that we replace $y_1=f(x_0)$ which makes sense since if you evaluate a function at a point it gives the y co-ordinate...next we replace $x_1=x_0$ which is the x co-ordinate lastly we need the slope at the point we want the tangent curve to go through which is $x_0$ and to get the slope of $f(x)$ at $x-0$ we need to find $f'(x_0)$...giving us the adapattion of hte pointslope formula the tangent line equation $y-x_0=f'(x_0)(x-x_0)$ and that is how/why we use the tangent line...any more questions just ask

5. do i always use this formula whenever determining the equation of the tangent line? because i've been taught to only use "y = mx + b", but i guess the point slope and the slope formulas are basically the same thing...?

6. ## Well

Originally Posted by Macleef
do i always use this formula whenever determining the equation of the tangent line? because i've been taught to only use "y = mx + b", but i guess the point slope and the slope formulas are basically the same thing...?
Don't differ from your teacher's curriculum...but I would because if you think about it they give you a point and a derivative which to evaluate it and get a slope....therefore you have a slope and a point....haha point-slope was made for this situation

7. They are equivalent:

$y - y_{1} = m(x - x_{1})$

You use this when you have the slope but no points of the equation. If you are given a point $(x_{1},y_{1})$ you can substitute it above and arrive at your desired form of $y = mx + b$