♦Learn basic concepts about quadratic functions and their
graphs.
♦Complete the square and apply the vertex formula.
♦Graph a quadratic function by hand.
♦ Solve applications and model data.
Basic Concepts
REVIEW:A linear function can be written as f(x) = ax+ b(or f(x) = mx+ b).
The formula for a quadratic function is different
from that of a linear function because it contains an x2 term.
f(x) = 3x2+ 3x+ 5
g(x) = 5 −x2
Quadratic Function Properties
•The graph of a quadratic function is a parabola—a U shaped
graph that opens either upward or downward.
•A parabola opens upward if its leading coefficient a is positive and
opens downward if a is negative .
•The highest point on a parabola that opens downward and the lowest point on a
parabola that opens upward is called the vertex. (The graph of a parabola
changes shape at the vertex.)
•The vertical line passing through the vertex is called the axis of symmetry.
•The leading coefficient a controls the width of the parabola. Larger
values of |a| result in a narrower
parabola, and smaller values of |a| result in a wider parabola.
Examples of different parabolas
Demonstrate EXCEL file for Quadratic Functions
DEMO
Example
Use the graph of the quadratic function shown to de termine the sign of the
leading coefficient, its vertex, and the equation of the axis of symmetry.
Leading coefficient: The graph opens downward, so
the leading coefficient a is negative.
Vertex: The vertex is the highest point on the
graph and is located at (1, 3).
Axis of symmetry: Vertical line through the vertex
with equation x= 1.
The quadratic function f(x) =ax2+ bx+ c can be written in an alternate form
that relies on the vertex (h, k).
Example: f(x) = 3(x -4)2+ 6 is in vertex form with
vertex (h, k) = (4, 6).
What is the vertex of the parabola given by
f(x) = 7(x + 2)2–9 ?
Vertex = (-2,-9)
Demonstrate EXCEL file for Quadratic Functions in
Vertex Form
DEMO
Example: Convert the quadratic f(x) = 3(x + 2)2–8which
is in vertex form to the form f(x) = ax2+ b + c
Given formula f(x) = 3(x + 2)2–8
Expand the quantity squared. f(x) = 3(x2+ 4x + 4) -8
Multiply by the 3. f(x) = 3x2+ 12x + 12 -8
Simplfy . f(x) = 3x2+ 12x + 4
Example
Write the formula f(x) = x2+ 10x+ 23 in vertex form by completing the
square.
| y = x2 + 10x + 23 |
Given formula |
| y - 23 = x2 + 10x |
Add itionsubtraction-reveiw-and.html">Subtract 23 from each side . |
| y - 23 + 25 = x2 + 10x + 25 |
Add (10/2)2 = 25 to both sides. |
| y + 2 = (x + 5)2 |
Factor perfect square trinomial . |
| y = (x + 5)2 - 2 |
Subtract 2 form both sides.. |
| What is the vertex? |
Vertex is h= -5 k = -2. |
Example Find the vertex of the graph of
a= 1/2, b= −4, and c= 8
x-coordinate of vertex:
y-coordinate evaluate f(4):
The vertex is (4, 0).
Example:
Use the vertex formula to write f(x) = −3x2−3x+ 1 in vertex form, then graph
the parabola.
1.Begin by finding the vertex.
2. Compute k = f(h).
3. The vertex is
4. The vertex form is
5. To graph the parabola we use the value of coefficient
a, the vertex and the axis of symmetry together with a few points on either side
of the line of symmetry.
a = -3 so the parabola opens downward.
The vertex is
So the axis of symmetry is the vertical line x = -1/2.
Next determine a few points on either side of the line of
symmetry.
Based on the figure compute the y-coordinates of f(x) =
−3x2−3x+ 1 for points in the table below.
| x |
f(x) |
| -2 |
-5 |
| -1 |
1 |
| -1/2 |
7/6 |
| 0 |
1 |
| 1 |
-5 |
Plot these points.
Connect the points with a smooth curve.
