F(x)=4.5x2–3x+2



use the limit definition to find f ′(x).

use the alternate limit definition, limit as x approaches a of the quantity f of x minus f of a over the quantity x minus a with a = 2, to find the value of f ′(2).

use the shortcut rules to find f ′(x).

find where the tangent line to f(x) is horizontal.

write the equation of the tangent line at x = 2.



what i've done so far:
alternate limit definition says that f'(x) = lim x-> a [f(x) - f(a)]/(x-a)
putting f(x) in above definition
we get
lim x-> 2 [ 4.5x2-3x+2 - 4.5*22+3*2-2]/(x-2)
lim x-> 2 [4.5x2-3x-12]/(x-2)
lim x-> 2 4.5*[x+12/9]*[x-2] /(x-2) as -12/9 and 2 are roots of x2-(3/4.5)x - 12/4.5
4.5*10/3 =15

let me know if my work is correct, see attached for easier viewing.

Your work is confusing. please consider using latex (clicking the π looking sign at the bottom)

1. means the slope at that point
not exactly sure what the limit defintion because I learned the one you call the alternate limit definition
ah, I got it

so





now simply set h=0
f'(x)=9x-3

2. nuuuh, not again






nrrrgh
oh, it divides nicely because 4.5x²-3x-12=(x-2)(4.5x+6)
4.5x+6
evaluate for x=2
4.5(2)+6
9+6
15

3.
dy/dx x^m=mx^(m-1)
dy/dx (f(x)+g(x))=dy/dx f(x)+dy/dx g(x)
dy/dx c=0 where c is a constnat
those are my shortcut rules

dy/dx 4.5x^2=4.5(2)x^(2-1)=9x^1=9x
dy/dx -3x=-3(1)x^(1-1)=-3x^0=-3(1)=-3
dy/dx 2=0
f'(x)=9x-3

4.
horizontal means slope=0
f'(x)=0
9x-3=0
9x=3
x=1/3
at x=1/3
at the point (1/3,3/2)

5.
at x=2, the slope is 15
at x=2, y=14
use point slope form
y-y1=m(x-x1)
point is (x1,y1)
slope is m
y-14=15(x-2)
most of the time it's alright to leave it in this form

Ugucycucucuvuvjvxyxtxu