1. 3.3 Quadratic Functions Objectives:
Define three forms for quadratic functions.
Find the vertex and intercepts of a quadratic function and sketch its graph.
Convert one form of a quadratic function to another.

2. Parabolas
vertex
The graph of a quadratic function is called a parabola and it always:
Opens upward or downward
Has a vertex which is a maximum or minimum
Has exactly 1 y-intercept
Has either 0, 1, or 2 x-intercepts
Is symmetric through its vertex
x-intercepts
y-intercept
Line of Symmetry

3. The Three Forms of a Quadratic Function
Transformation Form (Vertex Form):
f(x) = a(x – h)2 + k
Polynomial Form (Standard Form):
f(x) = ax2 + bx + c
x-Intercept Form (Factored Form):
f(x) = a(x – s)(x – t)

4. Graphing from the Three Quadratic Forms
Name
Transformation
Polynomial
x-Intercept
Form
Vertex
x-Intercepts
y-Intercept

5. Example #1 Transformation Form f(x) = a(x – h)2 + k
Find the vertex and the x- and y-intercepts for the following function. Then graph the function.
Vertex: (h, k)
x-Intercept: Let f(x) = 0
y-Intercept: Let x = 0
Vertex: (2, 5)
x-intercept:
y-intercept:
The graph opens down because a is negative.

6. Example #2 Polynomial Form f(x) = ax2 + bx + c
Find the vertex and the x- and y-intercepts for the following function. Then graph the function.
The vertex can be found from polynomial form using:
The graph opens up because a is positive.

7. Example #2 Polynomial Form f(x) = ax2 + bx + c
Find the vertex and the x- and y-intercepts for the following function. Then graph the function.
x-Intercept: Let f(x) = 0
y-Intercept: c 5

8. Example #3 x-Intercept Form f(x) = a(x – s)(x – t)
Find the vertex and the x- and y-intercepts for the following function. Then graph the function.
Vertex:
Graph opens down because a is negative.

9. Example #3 x-Intercept Form f(x) = a(x – s)(x – t)
Find the vertex and the x- and y-intercepts for the following function. Then graph the function.
x-Intercept: s & t
y-Intercept: ast

10. Example #4 Changing to Polynomial and x-Intercept Form
Write the following functions in polynomial and x-intercept form, if possible.
Polynomial:
x-intercept:
Since the discriminant is negative there are no x-intercepts so the function cannon be written into x-intercept form.

11. Example #4 Changing to Polynomial and x-Intercept Form
Write the following functions in polynomial and x-intercept form, if possible.
Polynomial: Already in polynomial form.
x-intercept:

12. Example #4 Changing to Polynomial and x-Intercept Form
Write the following functions in polynomial and x-intercept form, if possible.
Polynomial:
x-intercept: Already in x-intercept form.

13. Example #5 Changing to Transformation Form
Write the following functions in transformation form:
To write into transformation form we must use completing the square.
First factor out the a term.
Then complete the square and add this value to the inside of the parentheses.
Because of the -5 out front we are subtracting 9/5 so we must add 9/5 to the outside to keep it balanced.

14. Example #5 Changing to Transformation Form
Write the following functions in transformation form:

15. Example #6 Maximum Area for a Fixed Perimeter
Find the dimensions of a rectangular parking lot that can be enclosed with 2400 feet of fence and that has the largest possible area.
The largest area would occur if the parking lot were a square with 600 ft on each side and an area of 360,000 square feet.

16. Example #6 Maximum Area for a Fixed Perimeter
Find the dimensions of a rectangular parking lot that can be enclosed with 2400 feet of fence and that has the largest possible area.
Let x = length & y = width
Perimeter = x + x + y + y = 2x + 2y
Area = xy
This is a parabola opening downward so the maximum area should occur at its vertex.

17. Example #7 Maximizing Profit
The owner of a concession stand sells 150 lunches per day at a price of $3 each. The cost to the owner is $2.25 per lunch. Each $0.25 price increase decreases sales by 30 lunches per day. What price should be charged to maximize profit?
x = number of $0.25 price increases
The profit on each lunch is then 0.25x +0.75
The number of lunches sold per day is 150 – 30x
Total Profit: P(x) = (0.25x )(150 – 30x)
$3 – $2.25 = $0.75 per lunch

18. Example #7 Maximizing Profit
The owner of a concession stand sells 150 lunches per day at a price of $3 each. The cost to the owner is $2.25 per lunch. Each $0.25 price increase decreases sales by 30 lunches per day. What price should be charged to maximize profit?
The maximum profit occurs with just one price increase with each lunch priced at $3.25.

19. Example #8 Number Problems
Find two numbers whose difference is 7 and whose product is a minimum.
Even though is problem is looking for a minimum, the function is still a parabola so the vertex will be the lowest point.
The two numbers are 3.5 and −3.5.