1. 1.1 Graphing Quadratic Functions (p. 249)
Definitions
Standard form of quad. function
Steps for graphing
Minimums and maximums

3. Quadratic Function
A function of the form y=ax2+bx+c where a≠0 making a u-shaped graph called a parabola.
Example quadratic equation:

4. Vertex- Axis of symmetry- The lowest or highest point of a parabola .
This is the maximum or
minimum of the graph.
Vertex
Axis of symmetry-
The vertical line through the vertex of the parabola.
Axis of
Symmetry

5. Standard Form Equation
y=ax2 + bx + c
If a is positive, u opens up
If a is negative, u opens down
The x-coordinate of the vertex is at
To find the y-coordinate of the vertex, plug the x-coordinate into the given eqn.
The axis of symmetry is the vertical line x=
Choose 2 x-values on either side of the vertex x-coordinate. Use the eqn to find the corresponding y-values.
Graph and label the 5 points and axis of symmetry on a coordinate plane. Connect the points with a smooth curve.

6. Example 1
Graph a function of the form y = ax2 Graph y = 2x 2. Compare the graph with the graph of y = x 2.
SOLUTION
STEP 1
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.

7. Graph a function of the form y = ax 2
STEP 4
Compare the graphs of y = 2x 2 and y = x 2.
Both open up and have the same vertex and
axis of symmetry. The graph of y = 2x 2 is
narrower than the graph of y = x 2.

8. Example 2: Graph a function of the form y = ax 2 + c
graph of y = x 2
SOLUTION
Make a table of values for y = – x 12
STEP 1
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.

9. Example 2 continued
STEP 4 x 12

10. Practice X 2 Y Make a table of values for y = – x 2 – 5. – 2 – 1 – 9
SOLUTION
STEP 1
Make a table of values for y = – x 2 – 5. X
– 2
– 1 2 Y
– 9
– 6
– 5
STEP 2
Plot the points from the table.
STEP 3
Draw a smooth curve through the points.
STEP 4
Compare the graphs of y = – x 2 – 5 and y = x 2.

11. Practice Answer
ANSWER
Same axis of symmetry, vertex is shifted down 5 units, and opens down

12. Example 3: Graph y=2x2-8x+6 Axis of symmetry is the vertical line x=2
Table of values for other points: x y
0 6
1 0
2 -2
3 0
4 6
* Graph!
a=2 Since a is positive the parabola will open up.
Vertex: use
b=-8 and a=2
Vertex is: (2,-2)
x=2

13. Now you try one! y=x2−2x−1 * Open up or down? * Vertex? * Axis of symmetry?

14. Graph the function. Label the vertex and axis of symmetry.
y = x 2 – 2x – 1
SOLUTION
STEP 1
Identify the coefficients of the function. The coefficients are a = 1, b = – 2, and c = – 1. Because a > 0, the parabola opens up.
STEP 2
Then find the y - coordinate of the vertex.
(– 2) (1) =
= 1
x =
b 2a –
Find the vertex. Calculate the x - coordinate.
y = 12 – 2 • = – 2

15. Practice answer So, the vertex is (1, – 2). Plot this point.
STEP 3
Draw the axis of symmetry x = 1.
STEP 4
Select the point to the right or the left of the axis of symmetry (right=2, left = 0) to find another point to plot. Then plot the symmetrical point.

16. Find the minimum or maximum value
Tell whether the function y = 3x 2 – 18x + 20 has a minimum value or a maximum value. Then find the minimum or maximum value.
SOLUTION
Because a > 0, the function has a minimum value To find it, calculate the coordinates of the vertex.
x = −
b 2a
= –
(– 18) 2a
= 3
y = 3(3)2 – 18(3) = –7
ANSWER
The minimum value is y = –7. You can check the answer on a graphing calculator.

17. Solve a multi-step problem
Go - Carts
A go-cart track has about 380 racers per week and charges each racer $35 to race. The owner estimates that there will be 20 more racers per week for every $1 reduction in the price
per racer. How can the owner of the go-cart track maximize weekly revenue ?

18. SOLUTION
STEP 1
Define the variables. Let x represent the price reduction and R(x) represent the weekly revenue.
STEP 2
Write a verbal model. Then write and simplify a quadratic function.
R(x) = 13, x – 380x – 20x 2
R(x) = – 20x x ,300
STEP 3
Find the coordinates (x, R(x)) of the vertex.
x = –
b 2a
= –
(– 20)
Find x - coordinate.
= 8
Evaluate R(8).
R(8) = – 20(8) (8) + 13,300 = 14,580

19. ANSWER
The vertex is (8, 14,580), which means the owner should reduce the price per racer by $8 to increase the weekly revenue to $14,580.

20. What If ? In Example 5, suppose each $1 reduction in the price per racer brings in 40 more racers per week. How can weekly revenue be maximized?
SOLUTION
Define the variables. Let x represent the price reduction and R(x) represent the weekly revenue.
STEP 1
STEP 2
Write a verbal model. Then write and simplify a quadratic function.
R(x) = – 20x x ,300

21. Find the coordinates (x, R(x)) of the vertex.
STEP 3
Find the coordinates (x, R(x)) of the vertex.
Find x - coordinate.
Then, evaluate R(12.75).
R(12.75) = – 40(12.75) (12.75) + 13,300 =
ANSWER
The vertex is (12.75, 19,802.5), which means the owner should reduce the price per racer by $12.75 to increase the weekly revenue to $19,

22. Assignment p. 6 8-16 even, 19 & 20, 22-30 even, 55 & 56
For graphing problems:
Does the function have a max or min?