Lesson 1-2

Report
Lesson 2.2, page 273
Quadratic Functions
Objectives
Recognize characteristics of parabolas.
Graph parabolas.
Determine a quadratic function’s minimum or maximum value.
Solve problems involving a quadratic function’s minimum or maximum value.
Vocabulary
• Quadratic Function: a function that can be
written in the form
f(x) = ax2 + bx + c, a ≠ 0.
• Standard form: f(x) = a(x - h)2 + k
• Parabola: graph of quadratic function; Ushaped curve; symmetric
• Symmetry: when folded, two sides match
exactly
f(x) = a(x - h)2 + k
• Vertex: the point (h,k); highest or lowest point of parabola;
on axis of symmetry
• a: describes steepness and direction of parabola
• Axis of symmetry: x = h; fold line that divides parabola into
two matching halves
• Minimum: if a > 0, parabola opens up; vertex is minimum
point or lowest point of parabola; k is the minimum value
• Maximum: If a < 0, parabola opens down; vertex the
maximum point or lowest point of parabola; k is the
maximum value
Minimum (or maximum) function value for a
quadratic occurs at the vertex.
• Identify the vertex of each graph . Find the
minimum or maximum value.
• A.
B.
y
y
10
10
x
-10
-10
10
x
-10
-10
10
Review
Solving Quadratic Equations
Example:
 Get in standard form.
 Factor.
(3x + 7)(x – 3) = 0

3x+7=0 or x–3 = 0
Set each factor equal to zero.
 Solve.
 Check.
3x2 – 2x = 21
3x2 – 2x – 21 = 0
x = -7/3 or x = 3
If no “B” term, can use square root
property to solve for zeros.
For example, if x2 = 16, then x = ±4.
• Solve f(x) = 1 – (x – 3)2.
When factoring or square root property won’t work,
you can always use the QUADRATIC FORMULA
x
•
•
•
•
b 
b  4 ac
2
2a
Get in standard form: ax2 + bx + c = 0
Substitute for a, b, and c in formula.
Solve.
Check.
This works every time!!!
Try these. Solve using the Quadratic formula.
• 3x2 - 2x - 1 = 0
x2 + 2x = -3
Graphing Quadratic Functions
f(x) = a(x - h)2 + k
1. Determine whether the parabola opens up
(a > 0) or down (a < 0).
2. Find the vertex, (h,k).
3. Find x-intercepts by solving f(x) = 0.
4. Find the y-intercept by computing f(0).
5. Plot the intercepts, vertex, and additional
points. Connect with a smooth curve.
Note: Use the axis of symmetry (x = h) to plot
additional points.
Check Point 1, page 276
• Graph f(x) = -(x – 1)2 + 4.
y
10
x
10
-10
-10
Check Point 2, page 276
• Graph f(x) = (x – 2)2 + 1.
y
10
x
-10
-10
10
f(x) = ax2 + bx + c
• If equation is not in standard form, you may have to complete
the square to determine the point (h,k).
2
f ( x)  2 x  4 x  3
Graph of
f ( x)  2 x  4 x  3
2
Finding the vertex when in ax2 + bx + c form.
Vertex =
 b
 b 
, f 


 2a  
 2a
Check Point 3, page 279
• Graph f(x) = -x2 + 4x +1. Identify domain &
range.
y
10
x
-10
10
-10
Re-Cap
Quadratic Equations & Functions
ax2 + bx + c = 0 or f(x) = ax2 + bx + c,
a  0
• “x” is squared.
• Graph would be a curve.
• Solutions are at the x-intercepts. The “zeros” are the
solutions or roots of the equation.
• To solve: factor & use zero product property, take square
root of both sides (if no B term), or use other methods.
See Example 4, page 279
Check Point 4, page 280: f(x) = 4x2 - 16x + 1000
a) Determine, without graphing, whether the function has a
minimum value or a maximum value.
b)
Find the minimum or maximum value and determine
where it occurs.
c)
Identify the function’s domain and range.
Example 1- Application Problem
A baseball player swings and hits a pop fly straight up in the
air to the catcher. The height of the baseball, in meters, t
seconds after the hit is given by the quadratic function
h ( t )   4 . 9 t  34 . 3t  1
2
How long does it take for the baseball to reach its maximum
height?
What is the maximum height obtained by the baseball?
Solution
Vertex:
x
b
2a

 34.3
2(  4.9)
 3.5 seconds
h  3.5    4.9  3.5   34.3(3.5)  1
2
 61.025 m eters
Example 2 - Application
A farmer has 600 feet of fencing to enclose a
rectangular field. If the field is next to a river,
so no fencing is needed on one side, find the
dimensions of the field that will maximize the
area. Find the maximum area.
Area = (600  2x)(x)
A(x) = 2x2 + 600x
Find Vertex – to find max.
Solution
x
600
2(  2)
= 150
y = 2(150)2 + 600(150)
= 45000
x
x
600 – 2x
Width
(x) = 150
Length(600-2x) = 300
Max area = 45000

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