# Math Help - Derivative and Tangent Line

1. ## Derivative and Tangent Line

Find the Derivative of 3x^2-4x+5

lim x->0 I got 3(x+Δx)^2-4(x+Δx)+5-(3x^2-4x+5)
Δx
lim x->0 = 3(x^2+2xΔx+Δx)-4x-4Δx+5-3x^2+4x-5
Δx
lim x->0 = 3x^2+6xΔx+3Δx^2-4x-4Δx+5-3x^2+4x-5
Δx
lim x->0 = 6xΔx+3Δx-4
Δx
lim x->0 = Δx(6x+3Δx-4)
Δx
lim x->0 = 6x=3(0)-4

6x-4

Find an Equation of the Tangent Line to the Graph Above at the given point (2,9)

f(x) = 3x^2-4x+5
f(x) = 6x-4
f(x) = 6(2)-4 = 12-4 = 8
y-8=8(x-2)
y=8x-8

Are these right? If not can anyone help me?

2. ## Re: Derivative and Tangent Line

Originally Posted by Deo3560
Find the Derivative of 3x^2-4x+5

lim x->0 I got 3(x+Δx)^2-4(x+Δx)+5-(3x^2-4x+5)
Δx
lim x->0 = 3(x^2+2xΔx+Δx)-4x-4Δx+5-3x^2+4x-5
Δx
lim x->0 = 3x^2+6xΔx+3Δx^2-4x-4Δx+5-3x^2+4x-5
Δx
lim x->0 = 6xΔx+3Δx-4
Δx
lim x->0 = Δx(6x+3Δx-4)
Δx
lim x->0 = 6x=3(0)-4

6x-4

Find an Equation of the Tangent Line to the Graph Above at the given point (2,9)

f(x) = 3x^2-4x+5
f(x) = 6x-4
f(x) = 6(2)-4 = 12-4 = 8
y-8=8(x-2)
y=8x-8

Are these right? If not can anyone help me?
$f(x) = 3x^2-4x+5$

$f'(x) = 6x-4$

$f'(2) = 8$

tangent line equation at the point $(2,9)$ ...

$y-9 = 8(x-2)$

thank you