Question

Find the equation of the tangent line to the given curve at the point (0,9). y=4x^2+8x+9

Answer 1

First step,
find the dy/dx @ x = 0

4x^2 + 8x + 9
dy/dx = 8x + 8

@ x= 0

8(0) + 8
8

Therefore, slope = 8

Answer 2

first we are going to find the derivative of our function which is y' = 8x+ 8 then we are going to let x=0 in our slope equation y' so y'=8 and that is our slope

and now we have everything we need to find the tangent line m=8 x=0 y=9

let us use this equation where (y -y1) =m(x-x1) and therefore y-9=8(x-0) which is y=8x + 9:)

Answer 3

Next,
Use the point given, slope and (y -y1) =m(x-x1) to work out.

Answer 4

thank you :) life savers lol