M3U6D6 Warm-up:

Find the zeroes:1. x2 – 6x – 16 = 0

2. 2x2 + 7x + 5 = 0

(x-8)(x+2)=0x-8=0 or x+2=0 x=8 or x=-2

(2x+5)(x+1)=02x+5=0 or x+1=0 2x=-5 or x=-1 x=-5/2

HW Check: Document camera

M1U6D6 Axis of Symmetry and Graphing Quadratics

Objective:

To explore and graph quadratic functions.

Quadratic functions of the forms y = ax2 and y = ax2 + c

DEFINITIONS….

Standard Form of a Quadratic Function: A quadratic function is a function that can be written in the form y = ax2 + bx + c, where a ≠ 0. This form is called the STANDARD FORM OF A QUADRATIC FUNCTION.

Parabola: a “U-Shaped curve”

Axis of symmetry: The fold or line that divides the parabola into two matching halves.

Exploring Quadratic Graphs

Vertex:

If a > 0 in If a < 0 in

y = ax2 + bx + c y = ax2 + bx + c

then then

vertex is a minimum vertex is a maximum

Exploring Quadratic Graphs

Identify the vertex of each graph. Tell whether it is a minimum or a

maximum.

Exploring Quadratic Graphs

a.

The vertex is (1, 2).

b.

The vertex is (2, –4).

It is a maximum. It is a minimum.

Use the graphs below. Order the quadratic functions

(x) = –x2, (x) = –3x2, and (x) = x2 from widest to narrowest graph.

So, the order from widest to narrowest is (x) = x2, (x) = –x2,

(x) = –3x2.

12

12

(x) = –x2 (x) = x2 12

Of the three graphs, (x) = x2 is the widest and (x) = –3x2 is the narrowest.

12

(x) = –3x2

Exploring Quadratic Graphs

Reminder…..

+a in y = ax2 + bx + c makes it open up.

-a in y = ax2 + bx + c makes it open down.

If |a| is “small” (i.e. between 0 and 1) the graph is wide.

If |a| is “big” (i.e. greater than 1) the graph is narrow.

PROPERTY:

Graph of a Quadratic Function: The graph of y = ax2 + bx + c, where a ≠ 0, has the line x = -b/2a as its axis of symmetry. The x-coordinate of the vertex is –b/2a.

Graph the function y = 2x2 + 4x – 3.

Step 1: Find the equation of the axis of symmetry and thecoordinates of the vertex.

Find the equation of the axis of symmetry.x =b

2a– =

–42(2) = – 1

The x-coordinate of the vertex is –1.

y = 2x2 + 4x – 3

y = 2(–1)2 + 4(–1) – 3

= –5

To find the y-coordinate of the vertex, substitute –1 for x.

The vertex is (–1, –5).

Step 2: Find two other points.

Use the y-intercept.

For x = 0, y = –3, so one point is (0, –3).

Choose a value for x on the same side of the vertex.

Let x = 1

y = 2(1)2 + 4(1) – 3  

= 3

For x = 1, y = 3, so another point is (1, 3).

Find the y-coordinate for x = 1.

Step 3: Reflect (0, –3) and (1, 3) across the axis of symmetry to get two more points.

Then draw the parabola.

Aerial fireworks carry “stars,” which are made of a sparkler-like material, upward, ignite them, and project them into the air in fireworks displays. Suppose a particular star is projected from an aerial firework at a starting height of 610 ft with an initial upward velocity of 88 ft/s. How long will it take for the star to reach its maximum height? How far above the ground will it be?

The equation h = –16t2 + 88t + 610 gives the height of the star h in feet at time t in seconds.

Step 2:  Find the h-coordinate of the vertex.h = –16(2.75)2 + 88(2.75) + 610 Substitute 2.75 for t.h = 731 Simplify using a calculator.

The maximum height of the star will be about 731 ft.

Step 1:  Find the x-coordinate of the vertex.

After 2.75 seconds, the star will be at its greatest height.

b2a– =

–882(–16) = 2.75

Classwork:M1U6D6 CW

Homework:M1U6D6 HW