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# Find the equation of the lines... watch

1. Find the equation of the lines that pass through point (-1,3) and are tangent to the curve

So far:

f(x) =

f'(x) =

Therefore, tangent is:

y -

Is my next step to expand the right hand side and move the left hand side to the right?

I don't get how I'm going to find what a is since there's a^3 and a^2...I was thinking factorise 'a' then that will leave me with a quadratic to factorise but it doesn't...so confused!

Thanks
2. I think you have generated a family of lines that are tangent to your curve, parameterised by a, the x-value of the point at which they are tangent.

At this stage, I'd plug in x=-1, y=3 to restrict the family to those that pass through (-1, 3) and then solve for a.
3. (Original post by gottastudy)
Therefore, tangent is:

y -

Is my next step to expand the right hand side and move the left hand side to the right?
I think you have an error in that it should be (x-a) on the right.

I'd next substitute the point (-1,3) into your equation and solve for a. It does work out.
4. I've got a = -3/2

Is that wrong?
5. (Original post by gottastudy)
I've got a = -3/2

Is that wrong?
That's one answer.
6. (Original post by ghostwalker)
That's one answer.
Is the other 0? Thanks!
7. (Original post by gottastudy)
Is the other 0? Thanks!

Yes.
Attached Images

8. (Original post by ghostwalker)
Yes.
So now I have the 2 values of a, do I substitute back into y - ?
9. (Original post by gottastudy)
So now I have the 2 values of a, do I substitute back into y - ?
Since that's the equation of the tangent, then yes; though see my previous comment regarding the sign.

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