MATH 31 LESSONS PreCalculus 8. Sketching Functions.

MATH 31 LESSONS

PreCalculus

8. Sketching Functions

A. Relations and Functions

Whenever one variable (y) is affected by another variable (x),

we say they form a relation.

A function is a special relation in which every x-value has

at most one y-value.

That is, if an x-value could have more than one y-value,

it is not a function.

All functions are relations, but not all relations are functions.

Vertical Line Test

If a relation is a function, then any vertical line can cross the graph at most once.

If the vertical line crosses the graph more than once,

then it is not a function.

e.g. Which of the following relations are functions?

This is a function.

The vertical lines never cross twice (or more).

This is not a function.

A vertical line can cross twice.

This is not a function.

A vertical line can cross twice.

This is a function.

The vertical lines never cross twice (or more).

B. Parabolas

A parabola is a function of the form

cbxaxy 2

Basic form

The basic (simplest) form

of a parabola is

2xy

Vertex: (0, 0)

Opens upward

2xy y

x

Standard form

The standard form (also called the completed-square form)

of a parabola is

or

khxay 2

2hxaky

Vertex: (h, k)

khxay 2y

x

k

h

V(h, k)

Vertex: (h, k)

Axis of symmetry: x = h

khxay 2y

x

x = h

V(h, k)

The axis of symmetry cuts the graph “in half”

Vertex: (h, k)

Axis of symmetry: x = h

Max / Min Value: y = k

khxay 2y

x

y = kV(h, k)

If the graph opens down, k is a maximum value.

If the graph opens up, k is a minimum value.

If a < 0, then the parabola

opens downward

e.g. y = - x2

y = -3 (x - 2)2 + 7

khxay 2y

x

If a > 0, then the parabola

opens upward

e.g. y = x2 + 11

y = 3 (x - 1)2 - 4

khxay 2y

x

khxay 2y

x

y = k

x = h

V(h, k)

Ex. 1 Find the vertex of

by completing the square.

Try this example on your own first.Then, check out the solution.

20243 2 xxy

20243 2 xxy

xxy 24320 2

Isolate the x-variables

20243 2 xxy

xxy 24320 2

xxy 8320 2

Factor out the coefficient of the squared term

20243 2 xxy

xxy 24320 2

xxy 8320 2

Determine the constant that is needed to make a perfect square, by squaring half of the middle (linear) term.

162

82

20243 2 xxy

xxy 24320 2

xxy 8320 2

168316320 2 xxy

Add the coefficient to both sides.

Don’t forget that the 16 is multiplied by 3 on the right side. You must do that on the left as well.

20243 2 xxy

xxy 24320 2

xxy 8320 2

168316320 2 xxy

24328 xyComplete the square

24328 xy

2843 2 xy

Put the parabola in standard form

Vertex: (-4, -28)

24328 xy

2843 2 xy

Ex. 2 Fully sketch

Identify key features of the graph, including vertex,

max/min, axis of symmetry, and intercepts.

Try this example on your own first.Then, check out the solution.

21102 xxy

Complete the square to put in standard form

21102 xxy

21102 xxy

xxy 1021 2

Isolate the x-variables

21102 xxy

xxy 1021 2

xxy 1021 2

Factor out the coefficient of the squared variable

21102 xxy

xxy 1021 2

xxy 1021 2

Find the constant required to make a perfect square:

252

102

21102 xxy

xxy 1021 2

xxy 1021 2

25102521 2 xxy

Add the constant to both sides.

Don’t forget that the 25 is actually negative, due to the coefficient in front of the brackets.

21102 xxy

xxy 1021 2

xxy 1021 2

25102521 2 xxy

254 xy

Complete the square

Vertex: (5, 4)

Opens downward (a < 0)

Axis of symmetry: x = 5

Max Value: y = 4

254 xy

45 2 xy Put in standard form.

Find intercepts

y-int: (x = 0)

21102 xxy

Find intercepts

y-int: (x = 0)

(0, -21)

21102 xxy

210100 2

21

x-int: (y = 0)

21102 xxy

21100 2 xx

x-int: (y = 0)

21102 xxy

21100 2 xx

021102 xx

x-int: (y = 0)

21102 xxy

21100 2 xx

021102 xx

073 xx

x-int: (y = 0)

(3, 0) (7, 0)

21102 xxy

21100 2 xx

021102 xx

073 xx

7,3x

Sketch

y

x

y = 4

x = 5

V(5, 4)

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