M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
Objectives
1.
2.
2
Graph quadratic functions of the form f(x) = a(x – h) + k.
2
Graph quadratic functions of the form f(x) = ax + bx + c.
Vocabulary
Prior Knowledge
Completing the square.
New Concepts
2
1. Graph quadratic functions of the a(x – h) + k.
Look at each of the following functions and their graphs that were generated on a graphing calculator.
Notice the change in the vertex.
Do you see a pattern?
Example 1:
2
Graph f(x) = x – 3. What is the vertex and line of symmetry?
The -3 moves the parabola down 3 units.
Vertex: (0, -3)
Axis of symmetry: x = 0
The axis of symmetry is always x = the x-coordinate of
the vertex.
These are vertical shifts. We begin with the basic parabola with vertex at the origin and shift it up or
2
down the y-axis based on the number after the x term.
Now look at each of these functions. How does the graph and vertex change?
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M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
These are horizontal shifts. Notice the graph always shifts the opposite direction from what you would
expect. The (x – 2) moves the parabola to the right 2. The (x + 5) moves the parabola to the left 5. The
graph moves to the number that will make the value in the parentheses equal 0.
Now put these shifts together.
Example 2:
2
Graph f(x) = (x – 2) + 3. What is the vertex and line of symmetry?
The +3 moves the parabola up 3 units.
The -2 moves the parabola to the right 2 units.
Vertex: (2, 3)
Axis of symmetry: x = 2
The axis of symmetry is always x = the x-coordinate of
the vertex.
Example 3:
2
Graph f(x) = (x + 1) + 2. What is the vertex and line of symmetry?
The +2 moves the parabola up 2 units.
The +1 moves the parabola to the left 1 unit.
Vertex: (-1, 2)
Axis of symmetry: x = -1
The axis of symmetry is always x = the x-coordinate of
the vertex.
2
Example 4: Graph f(x) = (x – 3) – 5. What is the vertex and line of symmetry?
The -5 moves the parabola down 5 units.
The -3 moves the parabola to the right 3 units.
Vertex: (3, -5)
Axis of symmetry: x = 3
The axis of symmetry is always x = the x-coordinate of
the vertex.
2
Not all parabolas have exactly the same shape as f(x) = x . Some are skinnier, some wider, some upside
2
down. The coefficient of the x term determines that. All of the functions we have looked at so far had 1
2
for the coefficient of x . Again, look at these graphs and see if you can see the pattern.
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M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
2
If the coefficient of the x term is larger than 1, the graph is narrower. If the coefficient is smaller than 1,
the graph is wider.
2
As mentioned earlier, notice the effect of a negative coefficient of the x term.
2
For a general quadratic function, f(x) = ax + bx + c,
if a is positive, the parabola opens upward.
if a is negative, the parabola opens downward.
2
To graph a quadratic function in the form f(x) = a(x – h) + k, find the vertex and axis of symmetry,
determine the direction the parabola opens and if the shape will be the same, wider or narrower than the
basic parabola. Then find and plot a couple of other points on either side of the axis of symmetry. The xand y-intercepts are good points to use.
2
Example 5: Graph f(x) = – (x + 1) + 4.
Vertex: (-1, 4)
Axis of symmetry: x = -1
The parabola opens downward.
The graph will be the same as the basic
parabola.
x-intercepts: (-3, 0) and (1, 0)
y-intercept (0, 3)
2
Example 6: Graph f(x) = -2(x + 1) – 3 .
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M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
Vertex: (-1, -3)
Axis of symmetry: x = -1
The parabola opens downward.
The graph will be narrower than the basic
parabola.
x-intercept: non-real solutions so instead
select an x value to the left and to the right of
the vertex and find the corresponding yvalue.
x = -2, y = -7 (-2, -7)
x = 1, y = -11 (1, -11)
y-intercept:
(0, -5)
2
Graph quadratic functions of the form f(x) = ax + bx + c.
2
It is quite easy to find the vertex of a parabola when the function is written in f(x) = a(x – h) + k form.
2
However, most of the time the function is written in standard form: f(x) = ax + bx + c. Then it is not quite
so easy to find the vertex.
2
There are two ways to find it: (1) rewrite the function in f(x) a(x – h) + k by completing the square or (2)
use the vertex formula.
Rewrite by completing the square.
Example 7: Find the vertex.
2
y = x + 4x – 9
Replace f(x) with y.
2
y + 9 = x + 4x
Move the constant to the other side. Remember the
coefficient of the squared term must be 1.
2
y + 9 + 4 = x + 4x + 4
y + 13 = (x + 2)
2
f(x) = x + 4x – 9
2
2
y = (x + 2) – 13
Complete the square on the right. Find half of the
coefficient of the x-term and square it. Add that amount to
both sides.
Factor the right side. The number in the binomial is the
number we found when we took half of the coefficient of
the x-term.
Isolate y.
Vertex (-2, -13)
Example 8: Find the vertex.
2
y = -3x + 6x + 5
2
y – 5 = -3x + 6x
2
y – 5 = -3(x – 2x)
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f(x) = -3x + 6x + 5
Replace f(x) with y.
Move the constant to the other side.
The coefficient of the squared term must be 1, so factor
out -3.
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M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
2
Complete the square on the right inside the
parentheses. Find half of the coefficient of the x-term
and square it. Add that amount inside the parentheses.
This amount must be multiplied by -3 so we have
changed the right side by -3. We must change the left
side by -3 also.
2
Factor the right side. The number in the binomial is the
number we found when we took half of the coefficient of
the x-term.
y – 5 – 3 = -3(x – 2x + 1)
y – 8 = -3(x – 1)
2
y = -3(x – 1) + 8
Isolate y.
Vertex (1, 8)
This becomes very cumbersome and it turns out there is a formula that can be used instead.
Use the formula.
b
. Once you know the value for x, you can find the y value
2a
of the vertex by evaluating the function for that x value. In other words, you find the value of f(x).
The x-coordinate of the vertex is equal to 
 b  b 
The vertex is  
, f 
  .
 2a  2a  
2
f(x) = x + 4x – 9
Example 9 (same as Example 7): Find the vertex.
a = 1, b = 4, c = -9
x  
4
2(1)
Identify a, b, and c.
The x-coordinate of the vertex is equal to 
b
.
2a
x  2
2
f(-2) = (-2) + 4(-2) – 9
Substitute the x-value into the function to find the ycoordinate of the vertex.
f(-2) = 4 – 8 – 9
f(-2) = -13
Vertex (-2, -13)
Example 10 (same as Example 8): Find the vertex.
a = -3, b = 6, c = 5
x  
6
2( 3)
2
f(x) = -3x + 6x + 5
Identify a, b, and c.
The x-coordinate of the vertex is equal to 
b
.
2a
x 1
2
f(1) = -3(1) + 6(1) + 5
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Substitute the x-value into the function to find the ycoordinate of the vertex.
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M098
Carson Elementary and Intermediate Algebra 3e
Section 11.4
f(1) = -3 + 6 + 5
f(1) = 8
Vertex (1, 8)
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