Lesson 2.2, page 273
Recognize characteristics of parabolas.
Determine a quadratic function’s minimum or maximum value.
Solve problems involving a quadratic function’s minimum or maximum value.
• 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
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
Minimum (or maximum) function value for a
quadratic occurs at the vertex.
• Identify the vertex of each graph . Find the
minimum or maximum value.
Solving Quadratic Equations
Get in standard form.
(3x + 7)(x – 3) = 0
3x+7=0 or x–3 = 0
Set each factor equal to zero.
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
b 4 ac
Get in standard form: ax2 + bx + c = 0
Substitute for a, b, and c in formula.
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
Check Point 1, page 276
• Graph f(x) = -(x – 1)2 + 4.
Check Point 2, page 276
• Graph f(x) = (x – 2)2 + 1.
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).
f ( x) 2 x 4 x 3
f ( x) 2 x 4 x 3
Finding the vertex when in ax2 + bx + c form.
Check Point 3, page 279
• Graph f(x) = -x2 + 4x +1. Identify domain &
Quadratic Equations & Functions
ax2 + bx + c = 0 or f(x) = ax2 + bx + c,
• “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.
Find the minimum or maximum value and determine
where it occurs.
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
How long does it take for the baseball to reach its maximum
What is the maximum height obtained by the baseball?
h 3.5 4.9 3.5 34.3(3.5) 1
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.
y = 2(150)2 + 600(150)
600 – 2x
(x) = 150
Length(600-2x) = 300
Max area = 45000