H Holybond Feb 2016 6 0 United States Feb 25, 2016 #1 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!

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 Jun 2008 16,217 6,765 North Texas Feb 25, 2016 #2 Find $f'(2)$ using the limit formula. Tangent line equation will be $y-(-4)=f'(2) \cdot (x-2)$

H Holybond Feb 2016 6 0 United States Feb 25, 2016 #3 Thanks for replying so quickly, but did the (x-2) come from the point slope formula?

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Feb 25, 2016 #4 Holybond said: Thanks for replying so quickly, but did the (x-2) come from the point slope formula? Click to expand... Yes ... $y-y_1=m(x-x_1)$, where $y_1=-4$, $m=f'(2)$, and $x_1=2$

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

P Plato MHF Helper Aug 2006 22,490 8,653 Feb 25, 2016 #5 Holybond said: Hey, I'm new. I'm having some trouble. I'm given the problem: y = 4x-3x^2 , (2,-4) Click to expand... 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 said: Hey, I'm new. I'm having some trouble. I'm given the problem: y = 4x-3x^2 , (2,-4) Click to expand... The tangent line at $(2,-4)$ is: $\large{\displaystyle y+4=\left[\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}\right](x-2)}$

H Holybond Feb 2016 6 0 United States Feb 25, 2016 #6 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.

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 Jun 2008 16,217 6,765 North Texas Feb 25, 2016 #7 $f(x)-f(2)=(4x-3x^2)-(-4)$ Try again ...

H Holybond Feb 2016 6 0 United States Feb 25, 2016 #8 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: Feb 25, 2016

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.

skeeter MHF Helper Jun 2008 16,217 6,765 North Texas Feb 25, 2016 #9 Not $x=-8$ ... the limit, $f'(2)=-8$

H Holybond Feb 2016 6 0 United States Feb 25, 2016 #10 Ah, okay. So then that means Y-Y1=M(x-x1) =Y-(-4)=-8(x-2).