Investigate what happens when the equation of a graph is y = 1 x 2 Copy and complete the table of values for x -values between - 4 and 4 . x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y

We have, y = 1 x 2 By putting x = - 4 y = 1 - 4 2 = 1 16 By putting x = - 3 y = 1 - 3 2 = 1 9 By putting x = - 2 y = 1 - 2 2 = 1 4 By putting x = - 1 y = 1 - 1 2 = 1 By putting x = - 1 2 y = 1 - 1 2 2 = 4 By putting x = 1 2 y = 1 1 2 2 = 4 By putting x = 1 y = 1 1 2 = 1 By putting x = 2 y = 1 2 2 = 1 4 By putting x = 3 y = 1 3 2 = 1 9 By putting x = 4 y = 1 4 2 = 1 16 Therefore, the complete table is: x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y 1 16 1 9 1 4 1 4 4 1 1 4 1 9 1 16

Investigate what happens when the equation of a graph is y = 1 x 2 x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y Plot the points to draw the graph.

We have, y = 1 x 2 By putting x = - 4 y = 1 - 4 2 = 1 16 By putting x = - 3 y = 1 - 3 2 = 1 9 By putting x = - 2 y = 1 - 2 2 = 1 4 By putting x = - 1 y = 1 - 1 2 = 1 By putting x = - 1 2 y = 1 - 1 2 2 = 4 By putting x = 1 2 y = 1 1 2 2 = 4 By putting x = 1 y = 1 1 2 = 1 By putting x = 2 y = 1 2 2 = 1 4 By putting x = 3 y = 1 3 2 = 1 9 By putting x = 4 y = 1 4 2 = 1 16 Therefore, the complete table is: x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y 1 16 1 9 1 4 1 4 4 1 1 4 1 9 1 16 We know that the equations in the form of y = a x (where a is a whole number) are called reciprocal equations And graphs of reciprocal equations are called hyperbolas. Therefore, the graph is:

Investigate what happens when the equation of a graph is y = 1 x 2 x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y How does your graph differ from the hyperbola?

We have, y = 1 x 2 By putting x = - 4 y = 1 - 4 2 = 1 16 By putting x = - 3 y = 1 - 3 2 = 1 9 By putting x = - 2 y = 1 - 2 2 = 1 4 By putting x = - 1 y = 1 - 1 2 = 1 By putting x = - 1 2 y = 1 - 1 2 2 = 4 By putting x = 1 2 y = 1 1 2 2 = 4 By putting x = 1 y = 1 1 2 = 1 By putting x = 2 y = 1 2 2 = 1 4 By putting x = 3 y = 1 3 2 = 1 9 By putting x = 4 y = 1 4 2 = 1 16 Then, the complete table is: x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y 1 16 1 9 1 4 1 4 4 1 1 4 1 9 1 16 We know that the equations in the form of y = a x (where a is a whole number) are called reciprocal equations And graphs of reciprocal equations are called hyperbolas. Then, the graph is: Therefore, Graph is still disjoint, but both curves are above the x -axis on opposite sides of the y -axis.

Investigate what happens when the equation of a graph is y = 1 x 2 x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y Why do we not use x = 0 in the table of values?

We have, y = 1 x 2 Then, the complete table is: x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y 1 16 1 9 1 4 1 4 4 1 1 4 1 9 1 16 We can not use x = 0 in the above table because division by 0 is meaningless. That is y = 1 0 2 = Not define

Investigate what happens when the equation of a graph is y = 1 x 2 x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y What are the asymptotes of the graph you have drawn?

We have, y = 1 x 2 Then, the complete table is: x - 4 - 3 - 2 - 1 - 1 2 1 2 1 2 3 4 y 1 16 1 9 1 4 1 4 4 1 1 4 1 9 1 16 We know that the equations in the form of y = a x (where a is a whole number) are called reciprocal equations And graphs of reciprocal equations are called hyperbolas. Then, the graph is: One asymptote is the y -axis (the line x = 0 ). The other is the line y = q . That is y = 0 Therefore, y = 0 (the x -axis) and x = 0 (the y -axis) are the asymptotes of the graph.

As with the hyperbola, the standard form y = 1 x 2 + q can be used to work out the asymptotes. Given y = 1 x 2 + 3 , what would the asymptotes be?

According to the question As with the hyperbola, the standard form y = 1 x 2 + q . Given y = 1 x 2 + 3 we have to find asymptotes. We know that, In the general equation y = a x 2 + q One asymptote is the y -axis (the line x = 0 ). The other is the line y = q . That is y = 3 Therefore, y = 3 and x = 0 (the y -axis) are the asymptotes.

Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4

Author:Karen Morrison & Nick Hamshaw

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4

Attempt the free practice questions from Exercise 4: Exercise 18.4 with hints and solutions to strengthen your understanding. Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition solutions are prepared by Experienced Embibe Experts.

Questions from Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4 with Hints & Solutions

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Investigate what happens when the equation of a graph is $y = \frac{1}{x^{2}}$

Copy and complete the table of values for $x$ -values between $- 4$ and $4$ .

$y = \frac{1}{{(2)}^{2}} = \frac{1}{4}$	$- 4$	$- 3$	$- 2$	$- 1$	$y = \frac{1}{{(3)}^{2}} = \frac{1}{9}$	$\frac{1}{2}$	$1$	$2$	$3$	$y = \frac{1}{{(4)}^{2}} = \frac{1}{16}$
$y$

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Investigate what happens when the equation of a graph is $y = \frac{1}{x^{2}}$

$x$	$- 4$	$\frac{1}{16}$	$- 2$	$- 1$	$- \frac{1}{2}$	$\frac{1}{2}$	$1$	$\frac{1}{4}$	$3$	$4$
$y$

Plot the points to draw the graph.

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Investigate what happens when the equation of a graph is $y = \frac{1}{x^{2}}$

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$- \frac{1}{2}$	$\frac{1}{2}$	$1$	$2$	$3$	$4$
$y$

How does your graph differ from the hyperbola?

EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Investigate what happens when the equation of a graph is $y = \frac{1}{x^{2}}$

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$- \frac{1}{2}$	$\frac{1}{2}$	$1$	$2$	$3$	$4$
$y$

Why do we not use $x = 0$ in the table of values?

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Investigate what happens when the equation of a graph is $y = \frac{1}{x^{2}}$

$x$	$- 4$	$- 3$	$- 2$	$- 1$	$- \frac{1}{2}$	$\frac{1}{2}$	$1$	$2$	$3$	$4$
$y$

What are the asymptotes of the graph you have drawn?

EASY

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

As with the hyperbola, the standard form $y = \frac{1}{x^{2}} + q$ can be used to work out the asymptotes. Given $y = \frac{1}{x^{2}} + 3$ , what would the asymptotes be?

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Use what you have learned in your investigation to sketch the graphs of: $y = - \frac{1}{x^{2}}$

MEDIUM

Upper Secondary: IGCSE

IMPORTANT

Cambridge IGCSE® Mathematics Core and Extended Coursebook Second Edition>Chapter 18 - Curved Graphs>Exercise 18.4>Q 4

Use what you have learned in your investigation to sketch the graphs of: $y = x^{- 2} + 2$

Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4

Karen Morrison Mathematics Solutions for Exercise - Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4

Questions from Karen Morrison and Nick Hamshaw Solutions for Exercise 4: Exercise 18.4 with Hints & Solutions

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