using scale of 2cm to 30cm for x axis and 2cm to 1 unit on the y axis draw the graph of y= 3 sin 2x - cos x

1

Find any positive solutions of the equation

x2 = 1x + x

by a graphical method.

Completing the table below provides the points needed to draw the graphs y = x2 and y = 1x + x.

x00.51.01.52.02.53.0x200.2512.2546.2591x + xInfinity2.522.172.52.93.33

Where necessary the values have been rounded to 2 decimal places.

The graph below shows y = x2 and y = 1x + x.

The curves intersect where x = 1.45 and so this is the solution of the equation.



2

The graph below represents the function

f(x) = x2 − 3x − 3



Use the graph to determine

Using the graph:

(a)

the value of f(x) when x = 2

f(2) = −5

(b)

the value of f(x) when x = −1.5

f(−1.5) ≈ 3.6

(c)

the value of x for which f(x) = 0

Intercepts with x-axis are x = −0.8, 3.8

(d)

the minimum value of f(x)

fmin = −5.3

(e)

the value of x at which f(x) is a minimum

x = 1.5

(f)

the solution of x2 − 3x − 3 = 5

x = −1.7 and 4.7

(g)

the interval on the domain for which f(x) is less than −3.

0 < x < 3

Note

The domain of a function is the values of x for which the function is defined. This is covered in Unit G4.

3

Given that y = 2x2 − 9x + 4

(a)

copy and complete the table below

x−2−10246y304022

Missing values:

f(−1) = 2 × (−1)2 − 9 × (−1) + 4 = 15

f(2) = 2 × 22 − 9 × 2 + 4 = −6

x−2−10246y30154−6022

(b)

using a scale of 1 cm to represent 1 unit on the x-axis and 2 cm to represent 5 units on the y-axis, draw the graph of y = 2x2 − 9x + 4 for −2 ≤ x ≤ 6



(c)

use your graph to solve the equation

2x2 − 9x + 4 = 15

Using the intersection of y = 15 with y = 2x2 − 9x + 4 gives estimates of the solution of

2x2 − 9x + 4 = 15

as x = −1 and x = 5.5 (see graph above).

4

(a)

The grid on the following page shows the line, l, which passes through the points Q (0, −1) and R (3, 2).

(i)

Determine the gradient of the line, l.

Gradient = 2 − (−1)3 − 0 = 33 = 1

(ii)

Write down the equation of the line, l.

y-intercept is −1, so equation of l is

y = −1 + x

(Alternative method: equation is of the form y = mx + c = x + c;

to pass through the point (0, −1), −1 = 0 + c⇒ c = −1⇒ y = x − 1)

(b)

The table below shows three of the values of f(x) = x2 − 4x + 3 for values of x from 0 to 4.

x01234f(x)3−10

(i)

Copy the table and insert the missing values of f(x).

x01234f(x)30−103

(ii)

On a copy of the grid below, draw the graph of f(x) = x2 − 4x + 3.





(iii)

Using the graphs, write down the coordinates of the points of intersection of the line, l, and the graph of f(x).

Points of intersection at (1, 0) and (4, 3).

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