Mathematics » Differential Calculus » Sketching Graphs

Intercepts

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 \((4;0)\).

[Attributions and Licenses]


This is a lesson from the tutorial, Differential Calculus and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts