The objective of this module is to investigate the translation of a quadratic function and review quadratic functions.
Let
f
f
be the function given by
fx=x2−6x+p
f
x
x
2
6
x
p
, where
p
p
is an arbitrary constant.
Does
f
f
have a maximum or a minimum value? Explain your answer. Find the maximum or minimum value of
x
x
in terms of
p
p
.
The function
f
f
has a minimum value because the coefficient for the quadratic term,
x2
x
2
, is positive so the graph of the function will open upward and have a minimum value at the vertex. The
x
x
coordinate of the minimum value is
-b2a=--62×1=3
b
2
a
-6
2
1
3
. The minimum value of the function is
f3=p−9
f
3
p
9
.
If the value of
p
p
is changed, explain how this will affect the maximum/minimum value of
f
f
.
The
x
x
coordinate of the minimum value of
f
f
will not change because it is a constant (3). The minimum value is (
-9+p
-9
p
). Adding a constant,
p
p
, to the function
f
f
is a vertical translation of the graph of the function so the
x
x
coordinate, which is horizontal, will not change but the minimum value of
f
f
will increase or decrease depending on the value of
p
p
.
Find the value(s) of
p
p
where the function
f
f
has exactly one root. Graph this situation.
First method: Complete the square for
f
f
to find
p
p
.
fx=x2−6x+9
f
x
x
2
6
x
9
so
fx=x−32
f
x
x
3
2
Second method: Set the discriminant equal to zero then solve for
p
p
.
b2−4ac=0
b
2
4
a
c
0
-62−4×1p=0
6
2
4
1
p
0
p=9
p
9
Find the value(s) of
p
p
where
f
f
has no real roots.
Set the discriminant less than zero then solve for
p
p
.
b2−4ac<0
b
2
4
a
c
0
-62−4×1p<0
6
2
4
1
p
0
p>9
p
9
Observe that if
p>9
p
9
then the graph of
f
f will be translated above the x-axis and will have no real roots.
If
p=5
p
5
, find the roots for
f
f
. Find the average rate of change for
f
f
over the
x
x
interval
≤4
≤x
≤7
4
x
7
. Graph this situation in red on the same graph as problem 3.
If
p=5
p
5
, then the roots for
f
f
are 1 and 5.
The average rate of change is 5.
f7−f47−4=12−-33=5
f
7
f
4
7
4
12
-3
3
5
If
p=1
p
1
, find the roots for
f
f
. Find the average rate of change for
f
f
over the
x
x
interval
≤4
≤x
≤7
4
x
7
. Graph this situation in blue on the same graph as problem 3.
If
p=1
p
1
, then the roots for
f
f
are
3+22
3
2
2
or
3−22
3
2
2
.
The average rate of change is 5.
f7−f47−4=8−-73=5
f
7
f
4
7
4
8
-7
3
5
The average rate of change for
f
f
over the
x
x
interval
≤4
≤x
≤7
4
x
7
in problems 5 and 6 is the same. Explain why this rate is the same. If you change the value of
p
p
will the average rate of change always be the same for
f
f
over this interval? Justify your answer algebraically.
The average rate of change is the same because the values for
f4
f
4
and
f7
f
7
are vertically translated when the value of
p
p
is changed, therefore the secant line through these points is translated vertically and the two secant lines will be parallel.
If you change the value of
p
p
the secant lines for the translations will be parallel, therefore they will have the same slope and the average rate of change for
f
f
will be the same over this interval of
x
x
.
f4=-8+p
f
4
-8
p
and
f7=7+p
f
7
7
p
f7−f47−4=7+p−-8+p3=153=5
f
7
f
4
7
4
7
p
-8
p
3
15
3
5
Find the intervals of
x
x
where
f
f
is increasing/decreasing. If the value of
p
p
is changed will these intervals change? Justify your answer.
f
f
is decreasing over the interval
-∞3
3
.
f
f
is increasing over the interval
3∞
3
.
These intervals will not change because adding a constant to the function
f
f
translates the graph of
f
f
vertically so the minimum value will change but the
x
x
value where the function changes from decreasing to increasing will not change.
