Lecture 4 - Curve Sketching,Indices and Standard Form
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon [email protected] [email protected]
INTO City/UEA London
Lecture 4 skills
I find the y-intercept, x-intercept(s) and stationary point(s) ofa quadratic or cubic curve
I find the second derivative of an equationI determine the nature of a stationary pointI sketch a quadratic curveI sketch a cubic curve
I apply the rules of indices to simplify expressionsI convert a number to and from standard form
Lecture 4 vocabulary
I interceptI stationary pointI horizontalI maximumI minimumI point of inflectionI nth root, n
√
I standard form
Different types of curves to sketchStraight lines y = mx + c
Quadratic curves y = ax2 + bx + c, where a 6= 0:
Cubic curves y = ax3 + bx2 + cx + d , where a 6= 0:
Sketching curves
When sketching a curve, there are 4 main details you need toshow:
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 (ishorizontal).
4. How the curve behaves for large positive and largenegative values of x.
Stationary points
I Consider a curve with equation y = f (x).I A stationary point of y = f (x) is a point (a, f (a)) where the
curve has gradient 0 (is horizontal).I To find a stationary point you first find dy
dx . You then solve
the equation dydx = 0. The solutions are the x-coordinates of
the stationary points.I To get the y-coordinates, substitute the x-coordinates into
f (x).
Example 1I Question: What is the stationary point of y = x2 − 4x + 7?
2 4
2
4
6
•
I Answer: We have dydx = 2x − 4 (the gradient function).
I Solve dydx = 0.
If 2x − 4 = 0 then x = 2. This is the x-coordinate of thestationary point.
I The y-coordinate is y = 22 − 4× 2 + 7 = 3.I The stationary point of y = x2 − 4x + 7 is (2,3).
Example 2I Question: What are the stationary points of
y = x3 + x2 − 2x?
−2 −1 1
−4−2
24
•
•
I Answer: We have dydx = 3x2 + 2x − 2.
I Solve dydx = 0. It follows 3x2 + 2x − 2 = 0. Using the
quadratic formula we get x = −1±√
73 . These are the
x-coordinates of the two stationary pointsx = 0.55,−1.21 to 2 d.p.
I The y-coordinates arey = (−1±
√7
3 )3 + (−1±√
73 )2 − 2× −1±
√7
3 .y = −0.63,2.11 to 2 d.p.
I 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
I Consider the graph y = f (x).I A stationary point with x-coordinate a is a minimum if for all
points b close to a, we have f (b) > f (a).I A stationary point with x-coordinate a is a maximum if for
all points b close to a, we have f (b) < f (a).I 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
I How do we tell when a stationary point is a minimum,maximum or point of inflection?
I We use the second derivative.
I The derivative dydx is a function of x. This means we may
differentiate it again - to get the second derivative.I For the second derivative we use the notation
d2ydx2 .
I We say ’d squared y by d x squared’
The second derivative - examples
I What is the second derivative of y = x2 − 4x + 7?I Answer: We have dy
dx = 2x − 4.
I Then d2ydx2 = 2.
I What is the second derivative of y = x3 + x2 − 2x?I Answer: We have dy
dx = 3x2 + 2x − 2.
I Then d2ydx2 = 6x + 2.
Determining the nature of stationary points
Consider a stationary point of y = f (x) with x-coordinate a.
I If d2ydx2 (a) > 0 then the stationary point is a minimum.
I If d2ydx2 (a) < 0 then the stationary point is a maximum.
I If d2ydx2 (a) = 0 then we cannot tell.
The nature of the stationary points - example
I What is the nature of the stationary points of the curve withequation y = x3 + x2 − 2x?
I Answer: We have dydx = 3x2 + 2x − 2.
I Then d2ydx2 = 6x + 2.
I We previously found that the stationary points of this curvewere (0.55,−0.63) and (−1.21,2.11) to 2 d.p. We onlyneed to use the x-coordinate to determine whether eachpoint is a maxmum or minimum.
I d2ydx2 (0.55) = 6× 0.55 + 2 = 5.3 > 0 so this is a minimum
I d2ydx2 (−1.21) = 6×−1.21 + 2 = −5 < 0 so this is amaximum
The nature of the stationary points - example
I What is the nature of the stationary point of y = x2 − 1?I Answer: We have dy
dx = 2x
I and d2ydx2 = 2.
I d2ydx2 > 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 ofeach one. Find dy
dx (the derivative or gradient function).
Solve dydx = 0. Find d2y
dx2 (the second derivative). Evaluated2ydx2 for each stationary point.
4. Take a large positive number m and calculate f (m) andf (−m) to see if they are positive or negative.
Curve sketching - example 1
I Sketch the graph y = x2 − 1.I Answer: y = x2 − 1 passes through the y-axis when x = 0.
So y = 02 − 1 = −1.I The graph passes through the x axis when y = 0. Solve
0 = x2 − 1. So x = −1,1.I To work out the stationary points we solve dy
dx = 0. We
have dydx = 2x so x = 0. We need to determine the nature
of the stationary point. Use the second derivative. We haved2ydx2 = 2. Substitute in the x-coordinate of the stationary
point d2ydx2 (0) = 2 > 0 so the stationary point is a minimum.
I When x = 100 the y-value is large and positive. Whenx = −100 the y-value is large and positive.
Sketch of example
Curve sketching - example 2
I Sketch the graph y = x3 + x2 − 2x.I Answer: y = x3 + x2 − 2x passes through the y-axis when
x = 0. So y = 03 + 02 − 2× 0 = 0.I 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.
I To work out the stationary points we solve dydx = 0. We
have dydx = 3x2 + 2x − 2 so x = −1±
√7
3 . We need todetermine the nature of the stationary point. Use thesecond derivative. We have d2y
dx2 = 6x + 2. Substitute in the
x-coordinate of the stationary point d2ydx2 (0.55) > 0 so this
stationary point is a minimum. For the otherd2ydx2 (−1.21) < 0 so this stationary point is a maximum.
I When x = 100 the y-value is large and positive. Whenx = −100 the y-value is large and negative.
Sketch of example
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 ortwo x-intercepts.
The gradient function is dydx = 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 threex-intercepts.
The gradient function is dydx = 3ax2 + 2bx + c
Indices (powers)
xn = x × x × . . .× x
(x multiplied by itself n times)
What is 12, 13, 1100 ?x2 x squaredx3 x cubedxn where n ≥ 4 x to the power n
√x = 2√
x the square root of x3√
x the cube root of xn√
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√
xxn/m = ( m
√x)n = m
√xn
x0 = 1
Examples
I Simplify x2 × x3
I Answer:x2 × x3 = x2+3
= x5
I Simplify 3x1/2 × 14x3/2
I Answer:
3x1/2 × 14x3/2 = 3× x1/2 × 1
4 × x3/2
= 3× 14 × x1/2 × x3/2
= 34x(1/2+3/2)
= 34x2
Standard form
I Standard form provides a means way to write large orsmall numbers in a short way.
I
A× 10n
Where 1 ≤ A < 10 and n is an integer (n can be 0).
Standard form
I If the number is large count the number of digits to theright of the first digit.
I Express 1500 in standard form:I Answer: 1.5× 103.I Express 2,670,000 in standard form:I Answer: 2.67× 106.I If the number is small count the number of digits to the left of
the first non-zero digit.I Express 0.0035 in standard form:I Answer: 3.5× 10−3.I Express 0.0000007614 in standard form:I 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 form501/2
1/4
5110110
Multiplying in standard form
I Multiplication
(A× 10n)× (B × 10m) = (A× B)× 10n+m
I Question: Express the following in standard form
(3× 104)× (2× 102)
.I Answer:
(3× 104)× (2× 102) = (3× 2)× 104+2
= 6× 106
Multiplying in standard formI Question: Express the following in standard form
(2× 105)× (1× 10−3)
I Answer:
(2× 105)× (1× 10−3) = (2× 1)× 105+(−3)
= 2× 102
I Question: Express the following in standard form
(6× 105)× (4× 108)
I Answer:
(6× 105)× (4× 108) = (6× 4)× 105+8
= 24× 1013
= 2.4× 1014
Dividing in standard formI Division
(A× 10n)÷ (B × 10m) = (A÷ B)× 10n−m
I Question: Express the following in standard form
(4× 106)÷ (2× 103)
I Answer:
(4× 106)÷ (2× 103) = (4÷ 2)× 106−3
= 2× 103
I Question: Express the following in standard form
(2× 105)÷ (8× 108)
I Answer:
(2× 105)÷ (8× 108) = (2÷ 8)× 105−8
= 0.25× 10−3
= 2.5× 10−4