C2 st lecture 4 handout

Lecture 4 - Curve Sketching,
Indices and Standard Form
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon Craik
l.stringer@uea.ac.uk s.craik@uea.ac.uk
INTO City/UEA London

Lecture 4 skills
ﬁnd the y-intercept, x-intercept(s) and stationary point(s) of
a quadratic or cubic curve
ﬁnd the second derivative of an equation
determine the nature of a stationary point
sketch a quadratic curve
sketch a cubic curve
apply the rules of indices to simplify expressions
convert a number to and from standard form

Lecture 4 vocabulary
intercept
stationary point
horizontal
maximum
minimum
point of inﬂection
nth root, n
√
standard form

Different types of curves to sketch
Straight lines y = mx + c
Quadratic curves y = ax2 + bx + c, where a = 0:
Cubic curves y = ax3 + bx2 + cx + d, where a = 0:

Sketching curves
When sketching a curve, there are 4 main details you need to
show:
1. The y-intercept. Where the curve crosses the y-axis.
2. The x-intercept(s). Where the curve crosses (or touches)
the x-axis.
3. The stationary point(s). Where the curve has gradient 0 (is
horizontal).
4. How the curve behaves for large positive and large
negative values of x.

Stationary points
Consider a curve with equation y = f(x).
A stationary point of y = f(x) is a point (a, f(a)) where the
curve has gradient 0 (is horizontal).
To find a stationary point you first find dy
dx . You then solve
the equation dy
dx = 0. The solutions are the x-coordinates of
the stationary points.
To get the y-coordinates, substitute the x-coordinates into
f(x).

Example 1
Question: What is the stationary point of y = x2 − 4x + 7?
2 4
2
4
6
•
Answer: We have dy
dx = 2x − 4 (the gradient function).
Solve dy
dx = 0.
If 2x − 4 = 0 then x = 2. This is the x-coordinate of the
stationary point.
The y-coordinate is y = 22 − 4 × 2 + 7 = 3.
The stationary point of y = x2 − 4x + 7 is (2, 3).

Example 2
Question: What are the stationary points of
y = x3 + x2 − 2x?
−2 −1 1
−4
−2
2
4
•
•
Answer: We have dy
dx = 3x2 + 2x − 2.
Solve dy
dx = 0. It follows 3x2 + 2x − 2 = 0. Using the
quadratic formula we get x = −1±
√
7
3 . These are the
x-coordinates of the two stationary points
x = 0.55, −1.21 to 2 d.p.
The y-coordinates are
y = (−1±
√
7
3 )3 + (−1±
√
7
3 )2 − 2 × −1±
√
7
3 .
y = −0.63, 2.11 to 2 d.p.
The stationary points are (0.55, −0.63) and (−1.21, 2.11).

The nature of stationary points
A stationary point is exactly one of the following:
1. Minimum
2. Maximum
3. Point of inﬂection
Consider the graph y = f(x).
A stationary point with x-coordinate a is a minimum if for all
points b close to a, we have f(b) > f(a).
A stationary point with x-coordinate a is a maximum if for
all points b close to a, we have f(b) < f(a).
If a stationary point is neither a maximum nor a minimum it
is a point of inﬂection.

Maxima and minima - example
−2 2 4
−200
−100
100
200
x
y

Point of inﬂection - example
−2 −1 1 2
−5
5
•
x
y
y = x3

Determining the nature of a stationary point
How do we tell when a stationary point is a minimum,
maximum or point of inﬂection?
We use the second derivative.
The derivative dy
dx is a function of x. This means we may
differentiate it again - to get the second derivative.
For the second derivative we use the notation
d2y
dx2
.
We say ’d squared y by d x squared’

The second derivative - examples
What is the second derivative of y = x2 − 4x + 7?
Answer: We have dy
dx = 2x − 4.
Then d2y
dx2 = 2.
What is the second derivative of y = x3 + x2 − 2x?
Answer: We have dy
dx = 3x2 + 2x − 2.
Then d2y
dx2 = 6x + 2.

Determining the nature of stationary points
Consider a stationary point of y = f(x) with x-coordinate a.
If d2y
dx2 (a) > 0 then the stationary point is a minimum.
If d2y
dx2 (a) < 0 then the stationary point is a maximum.
If d2y
dx2 (a) = 0 then we cannot tell.

The nature of the stationary points - example
What is the nature of the stationary points of the curve with
equation y = x3 + x2 − 2x?
Answer: We have dy
dx = 3x2 + 2x − 2.
Then d2y
dx2 = 6x + 2.
We previously found that the stationary points of this curve
were (0.55, −0.63) and (−1.21, 2.11) to 2 d.p. We only
need to use the x-coordinate to determine whether each
point is a maxmum or minimum.
d2y
dx2 (0.55) = 6 × 0.55 + 2 = 5.3 > 0 so this is a minimum
d2y
dx2 (−1.21) = 6 × −1.21 + 2 = −5 < 0 so this is a
maximum

The nature of the stationary points - example
What is the nature of the stationary point of y = x2 − 1?
Answer: We have dy
dx = 2x
and d2y
dx2 = 2.
d2y
dx2 > 0, so this stationary point is a minimum.

How to sketch a curve y = f(x)
1. Find the y-intercept. Let x = 0. Evaluate y = f(0).
2. Find the x-intercept(s). Let y = 0. Solve f(x) = 0.
3. Find the stationary point(s), and determine the nature of
each one. Find dy
dx (the derivative or gradient function).
Solve dy
dx = 0. Find d2y
dx2 (the second derivative). Evaluate
d2y
dx2 for each stationary point.
4. Take a large positive number m and calculate f(m) and
f(−m) to see if they are positive or negative.

Curve sketching - example 1
Sketch the graph y = x2 − 1.
Answer: y = x2 − 1 passes through the y-axis when x = 0.
So y = 02 − 1 = −1.
The graph passes through the x axis when y = 0. Solve
0 = x2 − 1. So x = −1, 1.
To work out the stationary points we solve dy
dx = 0. We
have dy
dx = 2x so x = 0. We need to determine the nature
of the stationary point. Use the second derivative. We have
d2y
dx2 = 2. Substitute in the x-coordinate of the stationary
point d2y
dx2 (0) = 2 > 0 so the stationary point is a minimum.
When x = 100 the y-value is large and positive. When
x = −100 the y-value is large and positive.

Curve sketching - example 2
Sketch the graph y = x3 + x2 − 2x.
Answer: y = x3 + x2 − 2x passes through the y-axis when
x = 0. So y = 03 + 02 − 2 × 0 = 0.
The graph passes through the x axis when y = 0. Solve
0 = x3 + x2 − 2x. So 0 = x(x2 + x − 2) = x(x + 2)(x − 1).
It follows x = −2, 0, 1.
To work out the stationary points we solve dy
dx = 0. We
have dy
dx = 3x2 + 2x − 2 so x = −1±
√
7
3 . We need to
determine the nature of the stationary point. Use the
second derivative. We have d2y
dx2 = 6x + 2. Substitute in the
x-coordinate of the stationary point d2y
dx2 (0.55) > 0 so this
stationary point is a minimum. For the other
d2y
dx2 (−1.21) < 0 so this stationary point is a maximum.
When x = 100 the y-value is large and positive. When
x = −100 the y-value is large and negative.

Straight lines, y = mx + c
A straight line has gradient m.
It has exactly one x-intercept and one y-intercept.

Quadratic curves, y = ax2
+ bx + c
A quadratic curve has exactly one stationary point.
A quadratic curve has exactly one y-intercept and zero, one or
two x-intercepts.
The gradient function is dy
dx = 2ax + b

Cubic curves, y = ax3
+ bx2
+ cx + d
A cubic curve has zero, one or two stationary points.
A cubic curve has exactly one y-intercept and one, two or three
x-intercepts.
The gradient function is dy
dx = 3ax2 + 2bx + c

Indices (powers)
xn
= x × x × . . . × x
(x multiplied by itself n times)
What is 12, 13, 1100 ?
x2 x squared
x3 x cubed
xn where n ≥ 4 x to the power n
√
x = 2
√
x the square root of x
3
√
x the cube root of x
n
√
x where n ≥ 4 the nth root of x

Indices
xn
= x × x × . . . × x
x−1
= 1/x
x−n
= 1/xn
= 1/x × x × . . . × x
x0
= 1
What is 2−1 ?
What is 2−2, 3−3, 1−50 ?
What is 20, 30, 10, 10000, 00 ?

Indices
xn × xm = xn+m
xn ÷ xm = xn
/xm
= xn × (1/xm
) = xn × x−m = xn−m
(xn)m = xnm

Indices
x1/n
= n
√
x
xn/m
= (x1/m
)n = ( m
√
x)n
xn/m
= (xn)1/m
= m
√
xn

Indices - summary
m, n are real numbers and x can be a number or a variable
x−1 = 1/x
x−n = 1/xn
xn × xm = xn+m
xn ÷ xm = xn
/xm
= xn × (1/xm
) = xn × x−m = xn−m
(xn)m = xnm
x1/n
= n
√
x
xn/m
= ( m
√
x)n = m
√
xn
x0 = 1

Examples
Simplify x2 × x3
Answer:
x2 × x3 = x2+3
= x5
Simplify 3x1/2
× 1
4 x3/2
Answer:
3x1/2
× 1
4x3/2
= 3 × x1/2
× 1
4 × x3/2
= 3 × 1
4 × x1/2
× x3/2
= 3
4 x(1/2+3/2)
= 3
4 x2

Standard form
Standard form provides a means way to write large or
small numbers in a short way.
A × 10n
Where 1 ≤ A < 10 and n is an integer (n can be 0).

Standard form
If the number is large count the number of digits to the
right of the ﬁrst digit.
Express 1500 in standard form:
Answer: 1.5 × 103.
Express 2,670,000 in standard form:
Answer: 2.67 × 106.
If the number is small count the number of digits to the left of
the ﬁrst non-zero digit.
Express 0.0035 in standard form:
Answer: 3.5 × 10−3.
Express 0.0000007614 in standard form:
Answer: 7.614 × 10−7.

More practise with standard form
A × 10n
where 1 ≤ A < 10 and n is an integer (n can be 0)
Express the following in standard form
50
1/2
1/4
5
1
10
11
0

Multiplying in standard form
Multiplication
(A × 10n
) × (B × 10m
) = (A × B) × 10n+m
Question: Express the following in standard form
(3 × 104
) × (2 × 102
)
.
Answer:
(3 × 104) × (2 × 102) = (3 × 2) × 104+2
= 6 × 106

Multiplying in standard form
(2 × 105
) × (1 × 10−3
)
Answer:
(2 × 105) × (1 × 10−3) = (2 × 1) × 105+(−3)
= 2 × 102
(6 × 105
) × (4 × 108
)
Answer:
(6 × 105) × (4 × 108) = (6 × 4) × 105+8
= 24 × 1013
= 2.4 × 1014

Dividing in standard form
Division
(A × 10n
) ÷ (B × 10m
) = (A ÷ B) × 10n−m
(4 × 106
) ÷ (2 × 103
)
Answer:
(4 × 106) ÷ (2 × 103) = (4 ÷ 2) × 106−3
= 2 × 103
(2 × 105
) ÷ (8 × 108
)
Answer:
(2 × 105) ÷ (8 × 108) = (2 ÷ 8) × 105−8
= 0.25 × 10−3
= 2.5 × 10−4

C2 st lecture 4 handout

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C2 st lecture 4 handout