1. ## Another Polynomial question

The curve y=(ax+b)^2(c-x) cuts the x-axis at 3 and touches the x axis at x=9/4. Find the values of a,b,c.

2. Hello,
Originally Posted by requal
The curve y=(ax+b)^2(c-x) cuts the x-axis at 3 and touches the x axis at x=9/4. Find the values of a,b,c.
If one point is on the x-axis, then y=0. So the point (3,0) is a point on the curve.

If it "touches" the x axis at x=9/4, it means that the x axis is tangent to the curve at this point.
This means two things :
- the derivative at this point is 0.
- the point (9/4,0) is a point on a curve.

Now, substitute all this information in the equation. It should give you 3 equations with 3 unknowns.

3. Originally Posted by Moo
Hello,

If one point is on the x-axis, then y=0. So the point (3,0) is a point on the curve.

If it "touches" the x axis at x=9/4, it means that the x axis is tangent to the curve at this point.
This means two things :
- the derivative at this point is 0. Mr F adds: If you haven't done calculus, this means that there's a turning point at this point.

- the point (9/4,0) is a point on a curve.

Mr F adds: In case you haven't done calculus ..... when a cubic function has a turning point at the x-intercept ( ${\color{red}\alpha}$, 0) this means that ${\color{red}(x - \alpha)^2}$ is a factor of the cubic.

Now, substitute all this information in the equation. It should give you 3 equations with 3 unknowns.
..

4. Could you also take advantage of the fact that where it "touches" the x axis the power of the bracket must be even?

eg.

Since it cuts the x axis the power of the bracket must be odd (in this case 1):

I haven't worked through the rest of it yet but will this way work as well?

(you would need to find another equation linking and c and b or c and a. Differentiation can be used for this).