# Finding Equation of a Tangent Line without using Derivatives.

#### Holybond

Hey, I'm new. I'm having some trouble.

I'm given the problem: y = 4x-3x^2 , (2,-4)

I'm unsure about what to do exactly.

When do I use the slope of a line formula?

When am I supposed to use point slope?

When do I utilize the alternative
?

Please, and thanks for the help!

#### skeeter

MHF Helper
Find $f'(2)$ using the limit formula.

Tangent line equation will be

$y-(-4)=f'(2) \cdot (x-2)$

#### Holybond

Thanks for replying so quickly, but did the (x-2) come from the point slope formula?

#### skeeter

MHF Helper
Thanks for replying so quickly, but did the (x-2) come from the point slope formula?
Yes ... $y-y_1=m(x-x_1)$, where $y_1=-4$, $m=f'(2)$, and $x_1=2$

#### Plato

MHF Helper
Hey, I'm new. I'm having some trouble.
I'm given the problem: y = 4x-3x^2 , (2,-4)
The tangent line at $(2,-4)$ is:
$\large{\displaystyle y+4=\left[\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}\right](x-2)}$

#### Holybond

Thanks for the replies, but I'm a bit confused on both ends. F'(2) wouldn't I need to use the derivative to find that? And for Plato, when I do the math it should lead me to

(4x-3x^2)-4/x-2

= -(3x^2+4x+4)/X-2

Right? But that's not factorable.

#### skeeter

MHF Helper
$f(x)-f(2)=(4x-3x^2)-(-4)$

Try again ...

#### Holybond

Okay, I saw my mistake with the -(-4)

So this leads me to :

= -3x^2+4x+4/x-2

=-(3x^2 -4x-4)/x-2

= -(3x+2)(x-2)/x-2 They(x-2) cancel out.

= Leaving us with -(3x+2)

From there..I just plug the limit 2 into the problem which gives me. x=-8.

Last edited:

#### skeeter

MHF Helper
Not $x=-8$ ... the limit, $f'(2)=-8$

#### Holybond

Ah, okay. So then that means

Y-Y1=M(x-x1)

=Y-(-4)=-8(x-2).