Name:
Date:

The graph of any quadratic function f ( x ) = ax
2
+ bx + c is called a parabola .

The graph will have one of two shapes, and the a value tells which shape it will be. graph shape if a is positive graph shape if a is negative
Algebra 2

Every parabola has a special point called the vertex .
It’s the lowest or highest point.

Every parabola is symmetric across a vertical line called the axis of symmetry . The vertex is always on this line. The line’s equation is x = [the x -coordinate of the vertex].

Because of symmetry, when a quadratic function has two zeros, the vertex and the axis of symmetry are midway between the zeros. (In the example, the zeros are x = 3 and x = 7; the vertex and axis are at x = 5.)
Three ways to find the vertex
Here are three methods for finding the coordinates of the vertex, each covered by a part of today’s assignment:
(1) Calculator commands: [2nd]TRACE minimum or
[2nd][TRACE]maximum.
(2) x -coordinate of the vertex is the average of the zeros.
(3) formula for x -coordinate of the vertex: x =
b
.
2 a

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Date:
Algebra 2

First graph the function on the calculator ([Y=], enter formula, [GRAPH]).
 If you can’t see a 
or

shape, or the vertex isn’t on screen, press [WINDOW] and adjust.

Look at graph to see whether the vertex is the maximum (highest) or minimum (lowest) point.

Press [2nd][TRACE]maximum or [2nd][TRACE]minimum, whichever applies.

Move the cursor to the left of the vertex, then press [ENTER].

Move the cursor to the right of the vertex, then press [ENTER].

Press [ENTER] one last time, then the calculator displays the coordinates of the vertex.
1.
For each function, use your calculator to decide whether the vertex is a maximum or a minimum, find the coordinates of the vertex, and write an equation for the axis of symmetry. a.
f ( x ) = 2 x
2
+ 4 x + 7 maximum or minimum? vertex: axis of symmetry: b.
f ( x ) = –3 x
2
+ 6 x maximum or minimum? vertex: axis of symmetry: c.
f ( x ) = – x
2
+ 4 x + 10 maximum or minimum? vertex: axis of symmetry:
2.
Suppose that when a ball thrown from a cliff, its path is given by the equation: h ( x ) = – x
2
+ 10 x + 24. What is the maximum height reached by this ball?
3.
For the quadratic function f ( x ) = x 2 + 3 x – 24, make the table and graph on your calculator.
Then, use the calculator to find the zeros, vertex, and y -intercept. zeros : vertex : y intercept :

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Date:
Algebra 2
Here is how to use a quadratic’s zeros to find the coordinates of the vertex:

First find the zeros by any method (such as factoring or the Quadratic Formula).

Find the x -coordinate of the vertex by averaging the zeros (add the zeros then divide by 2).

Then, you can evaluate f ( x ) to find out the y -coordinate of the vertex.
Example: Find the vertex and the axis of symmetry of f ( x ) = x
2
+ 2 x – 35.
Solution:
Factor the function: f ( x ) = ( x – 5)( x + 7).
Then find the zeros: x = 5, x = –7 x -coordinate of vertex: x =
5
+
(
-
7 )
= –1.
2 y -coordinate of vertex: y = f (–1) = (–1)
2
+ 2(–1) – 35 = 1 – 2 – 35 = –36.
Answer: The vertex is (–1, –36). The axis of symmetry is the line x = –1.
4.
For each function, use factoring to find the zeros. Then, find the coordinates of the vertex and an equation for the axis of symmetry. a.
f ( x ) = x
2
– 4 x – 60 b.
f ( x ) = 6 x
2
– 5 x + 1 c.
f ( x ) = – x
2
+ 3 x + 70

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Date:
Algebra 2
5.
Find the zeros of these functions using factoring, then find the coordinates of the vertex. a.
f ( x ) = 5 x
2
+ 20 x + 15 b.
f ( x ) = 3 x
2 +
6 x c.
f ( x ) = 4 x
2
– 9
6.
Check your answers to all parts of problems 4 and 5 by finding the vertex of each function on your calculator. Record each vertex from the calculator here, and confirm agreement with your previous answers. Fix any mistakes that you find.
4a.
4b.
4c.
5a.
5b.
5c.

Name:
Date:
x
=
-
2 b a
Here is how to find a quadratic’s vertex using a formula.

The x -coordinate of a parabola’s vertex is always x =
b
2 a

Then, you can evaluate f ( x ) to find out the y -coordinate of the vertex.
Example: Find the vertex and the axis of symmetry of f ( x ) = –3 x
2
+ 12 x + 4.
Solution: x =
b
=
2 a 2
×
-
12
(
-
3 )
=
-
-
12
6
= 2. y = f (2) = –3 · 2
2
+ 12 · 2 + 4
= –12 + 24 + 4 = 16.
Answer: The vertex is (2, 16). The axis of symmetry is the line x = 2.
Algebra 2
7.
Using the formula shown above, find the vertex and the axis of symmetry for each of these functions. a.
f ( x ) = 5 x
2
– 20 x + 15 b.
f ( x ) = 3 x
2
+ 8 x + 6 c.
f ( x ) = x
2
– 7

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Date:
8 . For the following quadratic functions, find the vertex, zeros, axis of symmetry and y intercept. Use your information to sketch a graph of the function. a.
( ) = x
2 x
-
6
Vertex:__________________
Zeros:___________________
Axis of symmetry:_________
Y-intercept:______________
Algebra 2 b. f
=
6 x
x
2
Hint: First rearrange the terms into standard form.
Vertex:__________________
Zeros:___________________
Axis of symmetry:_________
Y-intercept:______________

Name:
Date:
9.
In St. Louis there is a large monument called the Gateway Arch that is shaped like a parabola. Specifically this function gives the shape of the arch is f ( x ) = –0.0089
x
2
+ 4.737
x .
The x -axis represents the ground, and both variables are measured in feet.
Algebra 2 a.
The arch is several hundred feet tall and several hundred feet wide. Press
[WINDOW] and choose Xmax and Gateway Arch photo: www.arrakeen.ch
Ymax that are large enough to see the whole shape of the arch. Record your choices.
Xmax = Ymax = b.
Graph the function on your calculator. Sketch the calculator screen in the box at the right. c.
Use the command [2nd][TRACE] Maximum to find the coordinates of the vertex. Write the coordinates of that point on your graph. d.
Use the command [2nd][TRACE] Zero twice to find the coordinates of the two zeros. Write the coordinates of those points on your graph. e.
How tall is the Gateway Arch, and how wide?
10.
In problem 7b , you found the vertex of f ( x ) = 3 x
2
+ 8 x + 6 using the formula method.
Explain why the method of averaging the zeros wouldn’t have worked for this function.
Hint: Look at the graph on your calculator.
11.
The function f ( x ) = x
2
+ 8 x + 16 has just one zero a.
Factor f ( x ), then find the zero of f ( x ). b.
Using any method, find the coordinates of the vertex of f ( x ).