Mathematics » Differential Calculus » Sketching Graphs

# Intercepts

## Optional Investigation: Number of intercepts

Complete the table:

 $$y = x + 1$$ $$y = x^{2} – x – 6$$ $$y = x^{3} + x^{2} – 26x + 24$$ Degree of function Type of function Factorised form No. of $$x$$-intercepts No. of $$y$$-intercepts

## Example

### Question

Given $$f(x)=-{x}^{3} + 4{x}^{2} + x – 4$$, find the $$x$$- and $$y$$-intercepts.

### Determine the $$y$$-intercept

The $$y$$-intercept is obtained by letting $$x = 0$$:

\begin{align*} y &= -(0)^{3} + 4(0)^{2} + (0) – 4 \\ &= -4 \end{align*}

This gives the point $$(0;-4)$$.

### Use the factor theorem to factorise the expression

We use the factor theorem to find a factor of $$f(x)$$ by trial and error:

\begin{align*} f(x) &= -x^{3} + 4x^{2} + x – 4 \\ f(1) &= -(1)^{3} + 4(1)^{2} + (1) – 4 \\ &= 0 \\ \therefore (x – 1) & \text{ is a factor of } f(x) \end{align*}

Factorise further by inspection:

\begin{align*} f(x) &= (x – 1)(-x^{2} + 3x + 4) \\ &= – (x – 1)(x^{2} – 3x – 4) \\ &= – (x – 1)(x + 1)(x – 4) \end{align*}

The $$x$$-intercepts are obtained by letting $$f(x) = 0$$:

\begin{align*} 0 &= – (x – 1)(x + 1)(x – 4) \\ \therefore x = -1, x = 1 & \text{ or } x = 4 \end{align*}

This gives the points $$(-1;0)$$, $$(1;0)$$ and $$(4;0)$$.