MATH 31 LESSONS PreCalculus 8. Sketching Functions.
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Transcript of 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)
3 7