Interpreting and Combining Graphs

Q: How do I find the range on a restricted domain?

Check the $y$-values at both endpoints and at any turning point inside the interval, then take the lowest and highest.

Q: How do I find the maximum distance between two graphs in Grade 11?

Write the distance as a quadratic in $x$ and find its turning point with $x = -\frac{b}{2a}$. In Grade 12 you can also use calculus.

Q: What does f(x) times g(x) greater than zero mean?

Both graphs are on the same side of the $x$-axis: both above or both below.

Many Grade 11 function questions give two graphs on the same axes and ask what they show together. You find where they meet, read off domain and range, measure distances between them and solve inequalities from the sketch.

Points of intersection

Set the two equations equal and solve. Then substitute to find $y$ .

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

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

The graphs meet at $(2; 3)$ and $(- 1; 0)$ .

Domain and range

Domain: the $x$ -values the function uses.
Range: the $y$ -values it produces.

Example. $f (x) = - x^{2} + 2 x + 3 = - (x - 1)^{2} + 4$ has domain $x \in R$ and range $y \leq 4$ .

Restricted domain. For $- 1 \leq x \leq 2$ , the highest point is the turning point $(1; 4)$ and the lowest is at $x = - 1$ , where $y = 0$ . So the range is $0 \leq y \leq 4$ .

When the domain is restricted, check the turning point and both endpoints.

Vertical and horizontal distances between graphs

Vertical distance at a given $x$ is top minus bottom:

$d (x) = top graph - bottom graph$

Example. Between $f$ (top) and $g$ for $- 1 \leq x \leq 2$ :

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

$d$ is a parabola with its maximum at $x = \frac{1}{2}$ : $d (\frac{1}{2}) = 2, 25$ units.

Horizontal distance is measured between $x$ -values at the same $y$ , such as the distance between the two $x$ -intercepts of a parabola.

Reading inequalities from graphs

$f (x) > g (x)$ : where $f$ is above $g$ .
$f (x) < 0$ : where $f$ is below the $x$ -axis.
$f (x) \cdot g (x) > 0$ : where both graphs are on the same side of the $x$ -axis.

Example. For $f$ and $g$ above:

$f (x) \geq g (x)$ for $- 1 \leq x \leq 2$
$f (x) \cdot g (x) > 0$ : both positive for $- 1 < x < 3$ , and both negative for $x < - 1$ . So $x < 3$ , $x \neq = - 1$ .

Use the intercepts and points of intersection as boundaries.

Key terms

Point of intersection: A point that lies on both graphs.
Restricted domain: A limited set of $x$ -values.
Vertical distance: The difference in $y$ -values at the same $x$ .

Common mistakes

Reading inequalities off the $y$ -axis instead of the $x$ -axis.
Forgetting the endpoints of a restricted domain when finding the range.
Subtracting bottom minus top for a distance.
Including $x$ -values where one graph is undefined.

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.

Worked solution

Hidden - try the question first

Solution

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

Question 2

4 marks total

routine

For $f (x) = - x^{2} + 2 x + 3$ :

(a) Write down the range of $f$ . (2)

(b) Write down the range of $f$ for $- 1 \leq x \leq 2$ . (2)

Worked solution

Hidden - try the question first

Solution

(a) $f (x) = - (x - 1)^{2} + 4$ , so the maximum is 4
$y \leq 4$
(b) $f (- 1) = 0$ and $f (2) = 3$ ; the maximum 4 is inside the interval
$0 \leq y \leq 4$

Question 3

6 marks total

complex

For $f (x) = - x^{2} + 2 x + 3$ and $g (x) = x + 1$ , determine the maximum vertical distance between $f$ and $g$ for $- 1 \leq x \leq 2$ .

Worked solution

Hidden - try the question first

Solution

$- x^{2} + 2 x + 3 = x + 1$
$= - x^{2} + x + 2$
$d$ is a parabola with maximum at $x = - \frac{b}{2 a} = \frac{1}{2}$
$d (\frac{1}{2}) = - \frac{1}{4} + \frac{1}{2} + 2$
$= 2, 25$ units

Question 4

8 marks total

problem-solving

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 $f (x) \cdot g (x) > 0$ ? (3)

(d) For which values of $x$ is $f$ increasing? (1)

Worked solution

Hidden - try the question first

Solution

(a) $f (x) = - (x - 1)^{2} + 4$
$\leq 2$
(b) Both positive: $- 1 < x < 3$
Both negative: $x < - 1$
$x < 3$ , $x \neq = - 1$
(c) The line $y = k$ must lie above the turning point
$k > 4$
(d) $x < 1$

FAQs

How do I find the range on a restricted domain?

Check the $y$ -values at both endpoints and at any turning point inside the interval, then take the lowest and highest.

How do I find the maximum distance between two graphs in Grade 11?

Write the distance as a quadratic in $x$ and find its turning point with $x = - \frac{b}{2 a}$ . In Grade 12 you can also use calculus.

What does f(x) times g(x) greater than zero mean?

Both graphs are on the same side of the $x$ -axis: both above or both below.

Points of intersection

Domain and range

Vertical and horizontal distances between graphs

Reading inequalities from graphs

Key terms

Common mistakes

Practice questions

FAQs

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