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Sketching graphs

Points of intersection

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Summary

Points of intersection

In a nutshell

Points of intersection are coordinates where two curves cross or touch each other. The $x$ coordinate of points of intersections can be found by solving the curves' equations as simultaneous equations. This can be done by equating the relevant equations and solving for $x$ .

Finding points of intersection

Points of intersection between two curves, $y=f(x)$ and $y=g(x)$ , can be estimated by sketching the curves on the same diagram. The exact coordinates can be found from the solution to $f(x)=g(x)$ .

Example 1

Sketch the functions $y=(x)(1-2x)$ and $y = (x)(x-3)(x+1)$ . How many points of intersection are there?

Identify the degree and sign of the polynomial, $y\text-intercept$ and the roots.

EQUATION	POLYNOMIAL DEGREE	SIGN	$y\textbf{\textit{-INTERCEPT}}$	ROOTS
$y=(x)(1-2x)$	$2$	$Negative$	$1$	$x = 0, \dfrac12$
$y = (x)(x-3)(x+1)$	$3$	$Positive$	$1 \times -3 = -3$	$x= -1,0, 3$

Use this information to sketch the curves.

Maths; Sketching graphs; KS5 Year 12; Points of intersection

Use the sketch to identify the number of points of intersection between the curves.

There are $\underline 3$ points of intersection.

Example 2

What are the coordinates of the points of intersection between the curves $f(x)=(x)^2(x-4)$ and $g(x) = (x)(2x-9)$ ?

Write $f(x)=g(x)$ in terms of $x$ .

$(x)^2(x-4) = (x)(2x-9)$ 

Expand the brackets.

$x^3-4x^2 = 2x^2 - 9x$ 

Collect the $x$ terms on one side.

$x^3 - 6x^2 +9x = 0$ 

Completely factorise the equation.

$\begin{aligned} &x(x^2 - 6x +9) &= 0 \\& x(x-3)^2 &=0 \end{aligned}$

Solve for $x$ .

$\begin{aligned} x&=0 \\ \\ x-3 &= 0 \\ x &=3 \end{aligned}$

Substitute $x$ into $f(x)$ or $g(x)$ to find the $y$ coordinate.

$\begin{aligned}g(0) &= (0)(2(0)-9) = 0 \\ g(3) &=(3)(2(3)-9)=(3)(-3) = -9 \end{aligned}$

Therefore, $f(x)$ and $g(x)$ intersect at the coordinates: $\underline{(0,0)}$ and $\underline{(3,-9)}$ .

Note: If an $x$ coordinate is a point of intersection, the $y$ coordinate will be the same if you substitute the $x$ value into the equation of either curve.

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