Maths (sketching a curve)

Sketching a curve using the following cubic function:
(just finding out the y-intercept and the x-intercept here)

$f(x)=(x-3)^2(x+1)$

Y-Intercept:

Let x = 0

$(x-3)^2(x+1)$

$(0-3)^2(0+1)$

$(-3)^2(1)$

$(-3)^2$

$9$

Y-Intercept = 9

X-Intercept:

Let f(x) = 0

$(x-3)^2(x+1)=0$

$(3-3)^2=0$

$(-1+1)=0$

X-Intercept = 3 or -1

Am I correct?

Reply 1

B_9710

You're all correct there as far as I can see, but it is worth noting the importance of the squared (x-3) term. This means that the x axis is a tangent to the curve at the root x=3. This is very important when sketching and although you could differentiate to find the turning points and see this anyway, it's much easier to just recognise it straight away. With this information, and the other root and the general shape of a cubic, you can't go far wrong.

Reply 2

Aklaol

Original post by B_9710

Yeah I know about the importance of the (x-3)^2. Since the cubic has a double root (which is x=3) it would mean when sketching the curve should only barely touch x=3, instead of going through it.

Reply 3

B_9710

Original post by Aklaol

Yeah I know about the importance of the (x-3)^2. Since the cubic has a double root (which is x=3) it would mean when sketching the curve should only barely touch x=3, instead of going through it.