# Thread: Equation of tangent line

1. ## Equation of tangent line

Hi!

Is my solution correct?

Let $\displaystyle f(x) = x^2 -1$. Find the equation of the line tangent to the graph of $\displaystyle f$ at the point where $\displaystyle x = 2$.

$\displaystyle \lim_{h \to 0} = \frac{f(x+h) - f(x)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{f(2+h)^2 - f(2)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{\left[(2+h)^2 - 1\right] - (2^2-1)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 4 - 4 + 1}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 1}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h(h+4) - 3}{h}$

$\displaystyle \lim_{h \to 0} (h+4) = -3$

Limit at $\displaystyle x=2$ doesn't exist.

2. ## Re: Equation of tangent line

Originally Posted by Unreal
$\displaystyle \lim_{h \to 0} (h+4) = -3$

Limit at $\displaystyle x=2$ doesn't exist.
That last line isn't correct and I'm not sure where it came from. Anyway...

I wouldn't have done it quite this way...the usual method is to plug in x after you get the expression for the derivative.
$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{ ( (x + h)^2 - 1 ) - (x^2 - 1)}{h}$

etc.
$\displaystyle f'(x) = \lim_{h \to 0} \frac{2hx + h^2}{h} = \lim_{h \to 0}(2x + h) = 2x$

and so forth. Can you take it from here??

-Dan

3. ## Re: Equation of tangent line

Originally Posted by Unreal
Hi!

Is my solution correct?

Let $\displaystyle f(x) = x^2 -1$. Find the equation of the line tangent to the graph of $\displaystyle f$ at the point where $\displaystyle x = 2$.

$\displaystyle \lim_{h \to 0} = \frac{f(x+h) - f(x)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{f(2+h)^2 - f(2)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{\left[(2+h)^2 - 1\right] - (2^2-1)}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 4 - 4 + 1}{h}$
This line is wrong. Your basic problem appears to be arithmetic.

$\displaystyle \lim_{h \to 0} = \frac{h^2 + 4h + 1}{h}$

$\displaystyle \lim_{h \to 0} = \frac{h(h+4) - 3}{h}$

$\displaystyle \lim_{h \to 0} (h+4) = -3$

Limit at $\displaystyle x=2$ doesn't exist.

4. ## Re: Equation of tangent line

Equation:

$\displaystyle y = m(x-a) + f(a)$

$\displaystyle y = 2(x-2) + 4$

OR

$\displaystyle y = 2x$

5. ## Re: Equation of tangent line

Originally Posted by Unreal
Equation:

$\displaystyle y = m(x-a) + f(a)$

$\displaystyle y = 2(x-2) + 4$

OR

$\displaystyle y = 2x$
f'(x) = 2x, so f'(2) = 2(2) = 4, not 2.

6. ## Re: Equation of tangent line

New equation,

$\displaystyle y = 4(x-2) + 4$

OR

$\displaystyle y = 4x -4, 4(x-1)$

7. ## Re: Equation of tangent line

Originally Posted by Unreal
New equation,

$\displaystyle y = 4(x-2) + 4$

OR

$\displaystyle y = 4x -4, 4(x-1)$
Still not going well. I don't think I've encountered that form for a line, or maybe I'm just not seeing it right. Try the slope-intercept form of a line:
y = mx + b

The slope for x = 2 is 2x = 2(2) = 4. So we have
y = 4x + b

You need to have the actual point where the tangent occurs, so what is f(2)? That will give you your point: (x, y) = (2, f(2)), and you plug that into your tangent line equation. That will give you the intercept for your tangent line.

By the way, recalculate f(2). It's not 4.

-Dan

8. ## Re: Equation of tangent line

$\displaystyle y = 3 + 4(x-2)$