Question

the curve y=ax^3 +bx passes throught the point (1,7). thr tangent at this point is parallel to the line y=2x-6

Answer 1

slope of tangent line is dy/dx
since it is parallel to y=2x-6, the slope must be 2

--> dy/dx = 2
--> 3ax^2+b = 2
x=1
--> 3a +b = 2

not sure what they want from here..

Answer 2

Not sure you even had to take the derivative because they give not only a slope but a point the line passes through.

Answer 3

oh right :{
so equation of tangent line is

y-7 = 2(x-1)

Answer 4

Yeah, I was waiting to see if sarah returned before answering, but she flew the coop as I was starting to reply and so I just stopped....lol

Answer 5

oh but if you need to solve for a,b you can use my prev post plus fact that if you plug in point you get

7 = a+b

Answer 6

I considered that but there was no directions when I got here, so I was going ask Sara what was the question wanting.?

Answer 7

hmm yes the entire question wasn't posted ... i usually speculate there are only so many unknowns to find :)

Answer 8

Ah well, guess she can post a reply if she returns.

Answer 9

I had a question the other day about finding the value of k in a problem that had no k to be found....after a series of questions, I was able to determine that they were comparing the equation to a standard form of a square root, and k was part of the vertex...lol.

Answer 10

hehe sorry they say find a and b

Answer 11

From what @dumbcow has shown, we get the system of equations:

3a + b = 2
a + b = 7

Now one can simply solve for a and b. I will use elimination to solve for a and b.

3a + b = 2
- a + b = 7
------------
2a = -5
a = -5/2

Now plug in this value of a back in to one of the equations and solve for b.

(-5/2) + b = 7
b = 7 + 5/2
b = 14/2 + 5/2
b = 19/2

Therefore, a = -5/2 and b = 19/2

@sarahlovesyumn

Answer 12

thank u :) but how did u get 3a+b=2?

Answer 13

@sarahlovesyumn, refer to very first post at top of page, i explain it there :)

Answer 14

to clarify more.... the derivative of function when x=1 is "3a+b"