Finding the Equation of a Tangent at a Given Point on a Cubic Curve

[TheRedDevil18]

Homework Statement

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

Homework Equations

y-y1=m(x-x1)

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?

 

[sankalpmittal]

Homework Statement

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

Homework Equations

y-y1=m(x-x1)

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?

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

 

[SteamKing]

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.

 

[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?

 

[SteamKing]

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.

 

[TheRedDevil18]

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

 

[HallsofIvy]

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

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

 

[SteamKing]

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

 

[TheRedDevil18]

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