Find an equation of the tangent line to y=(x^2)-3 at x=2.

Question

terenzreignz · Answer

Tangent lines... you mind should instantly jump to 'derivatives' :D
to get any line at all, you need its slope, and a point on that line.

The slope, you can find, by finding the derivative of this function... which is...?

anonymous · Answer

yup i'm doing derivatives in calc one right now. so to find the derivative I should plug in the two?

whpalmer4 · Answer

No, you find the derivative function with the power rule, then evaluate it at x = 2 to get the slope of the tangent line at x=2...

terenzreignz · Answer

First find the derivative.

whpalmer4 · Answer

Then you'll fit a line through the point (2,y(2)) with the slope you found.

whpalmer4 · Answer

Do you have the derivative yet?

whpalmer4 · Answer

Happy to help, but you need to do something...

anonymous · Answer

sorry I was doing something else but i'm back. ok so I'll try to find the derivative first

whpalmer4 · Answer

Good!

anonymous · Answer

ok 
so not quite sure if I did this right in finding the derivative

anonymous · Answer

$$x ^{2}+2x \Delta x-3+\Delta x ^{2}-3\div \Delta x$$

anonymous · Answer

that was by using the equation 
 = 	f(x+Δx) - f(x) / deltax

anonymous · Answer

i know I need to simplify it further... to 2xdeltax-6+deltax. but I thinking something is wrong there.

whpalmer4 · Answer

aren't you taking a limit?

anonymous · Answer

huh?

anonymous · Answer

cant you just use the power rule? $$nx^{n-1}$$

anonymous · Answer

or in your case $$ax^n =anx^{n-1}$$

whpalmer4 · Answer

what I was referring to is the definition of the derivative as

$$y'(x) = \lim_{\Delta x->0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

anonymous · Answer

ah i see will if limits are required by teacher to solving this answer ignore me :)

whpalmer4 · Answer

I would use the power rule, if you know it, but if you don't...

whpalmer4 · Answer

I doubt there's any requirement to do it the hard way, just that they may not have learned the easy way yet...

anonymous · Answer

ah i see well either way you get y prime so do what ever floats your boat :P

whpalmer4 · Answer

but here, we can do it the hard way, it isn't all that hard for this function!

Before we do, can you confirm that the equation is $$y = x^2-3$$ and not $$y = (x-3)^2$$

anonymous · Answer

yes its the first one.

anonymous · Answer

yup

whpalmer4 · Answer

$$y'(x) = \lim_{\Delta x->0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$$$y'(x) = \lim_{\Delta x->0}\frac{(x+\Delta x)^2-3 - (x^2-3)}{\Delta x}$$$$y'(x) = \lim_{\Delta x->0}\frac{(x^2+2x\Delta x + (\Delta x)^2-3 - (x^2-3)}{\Delta x}$$$$y'(x) = \lim_{\Delta x->0}\frac{(\cancel{x^2}+2x\Delta x + (\Delta x)^2\cancel{-3}\cancel{ - (x^2-3)}}{\Delta x} $$$$=\lim_{\Delta x->0} \frac{2x \Delta x + (\Delta x)^2}{\Delta x} = \lim_{\Delta x->0}(2x + \Delta x) = 2x$$

whpalmer4 · Answer

Okay, so the slope of the tangent to our curve at any point along the curve can be found by evaluating $y'(x) = 2x$.  We want to do this at $x = 2$.  

$y'(2) = 2(2) = 4$

So our tangent line has a slope of 4.

Now we use the point-slope formula for a line with known slope $m$ passing through a point $(x_0,y_0))$ to write the equation for the tangent line at $x=2$.

$$y-y_0 = m(x-x_0)$$
At $x=2$, $y = x^2 -3 = (2)^2-3 = 4-3 = 1$
Our known point is $(2,1)$

$$y - 1 = 4(x-2)$$$$y = 4x-8+1$$$$y = 4x-7$$is the equation of our tangent line at $x = 2$.

For your patience, a picture!

anonymous · Answer

wow your amazing! I was actually trying to work on it myself and that's the same thing I got! Thanks so much!

whpalmer4 · Answer

here's a question: what is the slope of the tangent line at x = 0?