Interpreting Graphs of Functions

Q: How do I read f(x) > g(x) from a graph?

Find where the graph of $f$ is above the graph of $g$, then give those $x$-values using the points of intersection as boundaries.

Q: What does f(x) times g(x) less than zero mean on a graph?

One graph is above the $x$-axis and the other is below it. Mark where each graph crosses the axis and test each region.

Many Paper 1 function questions give you two graphs on one set of axes and ask what they tell you. You need to find where they meet, how far apart they are, and for which $x$ -values one lies above the other.

Points of intersection

Two graphs meet where their $y$ -values are equal. Set $f (x) = g (x)$ , solve for $x$ , then substitute back to find $y$ .

Example. $f (x) = x^{2} - 2 x - 3$ and $g (x) = x + 1$ .

$x^{2} - 2 x - 3 = x + 1 \Rightarrow x^{2} - 3 x - 4 = 0 \Rightarrow (x - 4) (x + 1) = 0$

$x = 4$ gives $y = 5$ , and $x = - 1$ gives $y = 0$ . The graphs meet at $(4; 5)$ and $(- 1; 0)$ .

Substitute into the simpler function, usually the straight line, to find $y$ .

Vertical distances between graphs

The vertical distance between two graphs at a given $x$ is the top graph minus the bottom graph:

$d (x) = top - bottom$

To find the maximum vertical distance, differentiate $d (x)$ and set it equal to 0, or use the turning point of $d$ if it is a parabola.

Example. Between $g (x) = x + 1$ (top) and $f (x) = x^{2} - 2 x - 3$ for $- 1 \leq x \leq 4$ :

$d (x) = (x + 1) - (x^{2} - 2 x - 3) = - x^{2} + 3 x + 4$

$d^{'} (x) = - 2 x + 3 = 0 \Rightarrow x = 1, 5$

$d (1, 5) = 6, 25 units$

Check which graph is on top in the interval. If you subtract the wrong way, the distance comes out negative.

Reading inequalities from graphs

$f (x) > g (x)$ means the graph of $f$ is above the graph of $g$ . Read the $x$ -values from the points of intersection.

Example. With the graphs above, $f (x) \geq g (x)$ where the parabola is on or above the line: $x \leq - 1$ or $x \geq 4$ .

Other common forms:

$f (x) \cdot g (x) < 0$ : the graphs are on opposite sides of the $x$ -axis.
$\frac{f ( x )}{g ( x )} \geq 0$ : same sides, and $g (x) \neq = 0$ .
$f^{'} (x) > 0$ : $f$ is increasing.

Use strict inequalities ( $<$ , $>$ ) when the question is strict, and remember that $x$ -values where a function is undefined must be excluded.

Key terms

Point of intersection: A point that lies on both graphs, found by setting the equations equal.
Vertical distance: The difference in $y$ -values of two graphs at the same $x$ .
Critical values: The $x$ -values where the graphs meet or cross the axis; they divide the number line into regions to test.

Common mistakes

Finding only the $x$ -values of the intersection points when coordinates are asked for.
Subtracting bottom minus top for a vertical distance.
Reading inequalities from $y$ -values instead of $x$ -values.
Including asymptote values, such as $x = 2$ for a hyperbola, in a solution.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Given $f (x) = x^{2} - 2 x - 3$ and $g (x) = x + 1$ . Determine the coordinates of the points of intersection of $f$ and $g$ .

Worked solution

Hidden - try the question first

Solution

$x^{2} - 2 x - 3 = x + 1 \Rightarrow x^{2} - 3 x - 4 = 0$
$(x - 4) (x + 1) = 0$
$x = 4$ or $x = - 1$
$(4; 5)$ and $(- 1; 0)$

Question 2

5 marks total

routine

For the same $f$ and $g$ , determine the maximum vertical distance between the two graphs for $- 1 \leq x \leq 4$ .

Worked solution

Hidden - try the question first

Solution

$x^{2} - 2 x - 3 = x + 1 \Rightarrow x^{2} - 3 x - 4 = 0$
$= - x^{2} + 3 x + 4$
$d^{'} (x) = - 2 x + 3 = 0$
$x = 1, 5$
$d (1, 5) = - 2, 25 + 4, 5 + 4 = 6, 25$ units

Question 3

6 marks total

complex

Use $f (x) = x^{2} - 2 x - 3$ and $g (x) = x + 1$ .

(a) For which values of $x$ is $f (x) \geq g (x)$ ? (2)

(b) For which values of $x$ is $f (x) \cdot g (x) < 0$ ? (4)

Worked solution

Hidden - try the question first

Solution

(a) $x \leq - 1$
or $x \geq 4$
(b) $f$ has $x$ -intercepts $- 1$ and $3$ ; $g$ has $x$ -intercept $- 1$
For $x < - 1$ : $f > 0$ and $g < 0$ , so the product is negative
For $- 1 < x < 3$ : $f < 0$ and $g > 0$ , so the product is negative
$x < 3$ , $x \neq = - 1$

Question 4

9 marks total

problem-solving

Given $h (x) = \frac{4}{x - 2} + 1$ .

(a) Write down the equations of the asymptotes of $h$ . (1)

(b) The line $y = - x + k$ is an axis of symmetry of $h$ . Determine $k$ . (2)

(d) For which values of $x$ is $h (x) \geq x - 1$ ? (3)

Worked solution

Hidden - try the question first

Solution

(a) $x \leq - 1$ and $y = 1$
(b) The axes of symmetry pass through $(2; 1)$ : $1 = - 2 + k$
$k = 3$
(c) $(x - 4) (x + 1) = 0$
$(x - 2)^{2} = 4 \Rightarrow x - 2 = \pm 2$
$(4; 3)$ and $(0; - 1)$
(d) $h$ is on or above the line to the left of $(0; - 1)$ : $x \leq 0$
and between the vertical asymptote and $(4; 3)$ : $2 < x \leq 4$
$x = 2$ is excluded because $h$ is undefined there

FAQs

How do I read f(x) > g(x) from a graph?

Find where the graph of $f$ is above the graph of $g$ , then give those $x$ -values using the points of intersection as boundaries.

What does f(x) times g(x) less than zero mean on a graph?

One graph is above the $x$ -axis and the other is below it. Mark where each graph crosses the axis and test each region.

How do I find the maximum distance between two graphs?

Write the distance as top minus bottom, then find its maximum by differentiating or by using the turning point.

Points of intersection

Vertical distances between graphs

Reading inequalities from graphs

Key terms

Common mistakes

Practice questions

FAQs

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