Maths — Pure Mathematics 2

Can we sketch the graph $y=x^3-x^2-x+1$ ?

Review question2.1

Can we sketch the graph $y=x^3-x^2-x+1$ ?

Question

A sketch of the graph $y=x^3-x^2-x+1$ appears on which of the following axes?

Suggestion

What do we know about the shape of the curve from the coefficient of $x^3$ ?

Where does the curve cut the axes?

Can we factorise the polynomial expression?

Solution

The coefficient of $x^3$ is positive, so for large $x$ we get a positive $y$ , in which case (a) cannot be the graph.

When $x=0$ , $y=1$ , so the $y$ -intercept is $+1$ and therefore (b) cannot be the graph.

We note that if $x=1$ then $y=0$ , so the point $(1,0)$ is on the graph, which is not the case for (d).

Hence, the correct answer is (c).

Alternatively, we can factorise the graph into $y=(x+1)(x-1)^2$ , using the Factor Theorem.

This curve cuts the $x$ -axis at $-1$ , and touches it at $1$ . Hence, the correct answer is (c).

Can we sketch the graph

y=x^3-x^2-x+1

Question

A sketch of the graph

y=x^3-x^2-x+1

appears on which of the following axes?

Suggestion

What do we know about the shape of the curve from the coefficient of

x^3

Where does the curve cut the axes?

Can we factorise the polynomial expression?

Solution

The coefficient of

x^3

is positive, so for large

x

we get a positive

y

, in which case (a) cannot be the graph.

When

x=0

y=1

, so the

y

-intercept is

+1

and therefore (b) cannot be the graph.

We note that if

x=1

then

y=0

, so the point

(1,0)

is on the graph, which is not the case for (d).

Hence, the correct answer is (c).

Alternatively, we can factorise the graph into

y=(x+1)(x-1)^2

, using the Factor Theorem.

This curve cuts the

x

-axis at

-1

, and touches it at

1

. Hence, the correct answer is (c).