- #1
awholenumber
- 200
- 10
First of all there is an equation
Then there is the derivative
Then there is a point slope formula to find the equation of the tangent line
Point slope formula to obtain the tangent line .
y=3a2(x-a)+a3
Then Plug in the x coordinate into the derivative to get the slope
f'(1) = 3(1)2
f'(1) = 3
What this means is that for any value of x=a, the instantaneous slope of f at (a,a3) is 3a2.
Here , i don't really understand some change of terms from f(x) to y , a ... etc
Please help