# TANGENT???????????????????

The slope of the tangent line to the parabola y = 3 x^2 + 4 x + 2 at the point ( 2 , 22 ) is: 2

The equation of this tangent line can be written in the form y = mx+b where m is:_______________?

and where b is:_______________?

Relevance

Hi Fapre,

Agree with Iobo that m is the slope of the tangent line and b is the y-intercept of the tangent line. However, I would just add what the actual values are when applied to the parabola and its tangent you cited.

In this case, y' = 6x + 4 means that y'(2) = 16. So the the slope of the tangent line at x=2 is m=16.

To find b, the y-intercept of the tangent line, just use y=mx+b, your point (2,22) as (x,y), and m=16:

That gives 22 = 16(2) + b, or b = -10. Thus the equation of the tangent line to the parabola is y = 16x - 10.

To help reinforce your understanding of these concepts, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.

As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

• Anonymous
9 years ago

First, find the derivative:

dy/dx=6x+4

Now, plug in x:

Now plug this into the y=mx+b equation given your x,y coordinates to solve for b:

22=16(2)+b

22=32+b

b=-10

So:

m is 16

b is -10

• y = mx + b is known as the "slope/intercept" form of a line equation.

m is the slope of the line and b is the y intercept of the line.

You plug in the 3 known quantities x, y and m (all are given to you) and solve for b.

• y = 3x² + 4x + 2

dy/dx = 6x + 4

x = 2

y = 22

dy/dx = 6(2) + 4

dy/dx = 12 + 4

dy/dx = 16

b = y - [(dy/dx)(x)]

b = 22 - [16(2)

b = 22 - 32

b = - 10

Equation of Tangent Line:

y = 16x - 10

¯¯¯¯¯¯¯¯¯¯¯

m = 16

¯¯¯¯¯¯¯

b = - 10

¯¯¯¯¯¯¯

Source(s): 5/9/12