# Caculus f'(x)

posted by .

For what values of a and b is the line –2x+y=b tangent to the curve y=ax^3 when x=–2?

a= ?
b= ?

Thanks guys!

• Caculus f'(x) -

First, you rewrite the equation as:

-2∙-2 + a∙-2^3

Next, you solve some of it to get:

4 + a∙8

Finally, if you can't figure out what a and b are, you can do what I did above, and try figuring it out. I did not know how to solve to get a and b, but at least you have the things above to help.

• Calculus f'(x) (Misspelled Subject Name) -

Oh, and by the way, it is actually Calculus instead of Caculus for the subject name of your problem. I just noticed it.

• Caculus f'(x) -

For a if it is 4, you will get the next equation of:

4 + 4∙8 = b

Finally, you solve, and you get:

So, a equals 4 and b equals 36.

Hint: you can always estimate values for a and then you solve, and whatever you get as your answer will be b.

• Caculus f'(x) -

y = ax^3
y' = 3ax^2
at x=-2, y' = 12a
The line y=2x+b has slope 2, so
12a=2
a = 1/6

at x = -2, y = -8a = -4/3 so the line must go through (-2,-4/3)

So, the tangent line has slope 2 and goes through (-2,-4/3)

y = 2x+b
-4/3 = -4 + b
b = 8/3

So, now we have

y = 2x - 8/3
is tangent to
y = 1/6 x^3
at (-2,-4/3)

To confirm this, visit

http://www.wolframalpha.com/input/?i=plot+y+%3D+1%2F6+x^3+and+y+%3D+2x%2B8%2F3+for+x+%3D+-4+..+4

• Caculus f'(x) - typo -

typo near the end

y = 2x + 8/3

## Similar Questions

1. ### calc

for the parametric curve defined by x=3-2t^2 and y=5-2t ...sketch the curve using the parametric equation to plot of the point. use an arrow to indicate the direction of the curve for o<t<1. Find an equation for the line tangent …
2. ### Help Calc.!

original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point …

original curve: 2y^3+6(x^2)y-12x^2+6y=1 dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point …
4. ### math

With regards to question J: The variables x and y are connected by the equation y = x2 - x - 5. Some corresponding values of x and y are given in the table below. x -4 -3 -2 -1 0 1 2 3 4 5 y 15 7 a -3 -5 b -3 1 7 15 (a) Calculate the …
5. ### Pure Mathematics

1. Given that the straight line y=c-3x does not intersect the curve xy=3, find the range of values for c. 2. Find the range of values for c for which the line y=cx+6 does not meet the curve 2x^2-xy=3. 3. Find the range of values for …
6. ### Calculus

1. On what interval is the function f(x)=x^3-4x^2+5x concave upward?
7. ### Calc AB

Suppose that f(x) is an invertible function (that is, has an inverse function), and that the slope of the tangent line to the curve y = f(x) at the point (2, –4) is –0.2. Then: (Points : 1) A) The slope of the tangent line to the …
8. ### Maths

I need some help with my maths questions please! What angle does the tangent to the curve y=x^4-x^3+7x make with the positive direction of the x-axis at the point where x=2?
9. ### Calculus AB

Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x-1)e^x. For which value of x is the slope of the tangent line to f positive?