Tangent Lines: Understanding the equation

[James98765]

Hey Guys!
I have a problem with understanding the formula used to calculate tangent lines and was wondering if someone could explain it to me. Let me use an example problem to show what confuses me.

Problem:
Find the slope and equation of the tangent line to the graph of f(x)=x[sup:gguynkeq]2[/sup:gguynkeq] at point (3,9)

Step 1:
First I start with finding the slope of a secant line
m[sub:gguynkeq]sec[/sub:gguynkeq]=(f(3+h) - f(3))/h

Step 2:
Then I use the point in the formula
m[sub:gguynkeq]sec[/sub:gguynkeq]=(f(3+h) - f(3))/h

Step 3:
This step is what confuses me. My book says to square f(3+h) and f(3) like this:
m[sub:gguynkeq]sec[/sub:gguynkeq]=(f(3+h)[sup:gguynkeq]2[/sup:gguynkeq]-f(3)[sup:gguynkeq]2[/sup:gguynkeq])/h
The book says to do this because f(x)=x[sup:gguynkeq]2[/sup:gguynkeq]. My problem is I do not understand how the function of the graph would cause me to square something involving the slope of the secant line.
The rest of the steps I understand becuase I then just have to simplify the problem to find the slope of the secant and then get the tangent from that. If someone could explain the reasoning behind step 3 for me that would be great. Thanks!

 

[skeeter]

might be a notation error ...

\(\displaystyle f(3+h) = (3+h)^2 = 9 + 6h + h^2\)

\(\displaystyle f(3) = 9\)

\(\displaystyle f(3+h) - f(3) = 9 + 6h + h^2 - 9 = 6h+ h^2\)

\(\displaystyle \frac{6h+h^2}{h} = \frac{h(6 + h)}{h} = 6+h\)

\(\displaystyle \lim_{h \to 0} 6 + h = 6\)

slope is m = 6 ... now use the point-slope form to arrive at the tangent line equation.

 

[Dr. Flim-Flam]

f(x) = x^2, f ' (x ) = 2x = m, f ' (3) = 6

Now we have a slope of 6 and a point of (3,9),

hence y-9 = 6(x-3) , y = 6x-18+9 = 6x-9. QED

 

[James98765]

I got it. Thanks alot guys. I can see it now.