# Find Equation of Tangent

1. Mar 9, 2013

### TheRedDevil18

1. The problem statement, all variables and given/known data
The equation of the tangent to the curve f(x)=ax^3+bx at the point of contact (-1;3) is
y-x-4=0. Calculate the values of a and b

2. Relevant equations

y-y1=m(x-x1)

3. The attempt at a solution

I am totally stuck, here is what I could derive:
Equation of tangent y=x+4
f'(x)=3ax^2+b

Dont know what to do, can someone please help?

2. Mar 9, 2013

### sankalpmittal

Hint: Equation of tangent is given by :

y-y1=f'(x)*(x-x1)

You are already given x1 and y1. Compare this with the already given equation of tangent.

Note: f'(x)=(dy/dx) at x=x1 and y=y1

3. Mar 9, 2013

### SteamKing

Staff Emeritus
You have one point common to the curve and the tangent line.
You know the value of the slope at the tangent point.

You have two equations and two unknowns.

4. Mar 9, 2013

### TheRedDevil18

Ok, set up two equations:
a-b=3
3a+b=0
Make a subject of formula:
a=3+b

plug into equation 2 and solved, final answer:
b=-9/4
a=3/4

All good?

5. Mar 9, 2013

### SteamKing

Staff Emeritus
What is the slope of the tangent line at point (-1,3)?

Your second equation is incorrect.

Always check your solutions by substituting back into original equations.

6. Mar 10, 2013

### TheRedDevil18

The slope would be 1, because y=mx+c and equation is y=x+4

7. Mar 10, 2013

### HallsofIvy

Staff Emeritus
So why did you, previously, set 3a+b equal to 0?

8. Mar 10, 2013

### SteamKing

Staff Emeritus
How does slope = 1 affect your second equation for a and b?

9. Mar 10, 2013

### TheRedDevil18

Sorry guys, just slipped my mind that derivative is same as gradient, so my final answers should be:
a=1
b=-2

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted