10.2 Quadratic Functions. 10.2 – Quadratic Functions Goals / “I can…” I can graph quadratic...

10.2

Quadratic Functions

10.2 – Quadratic Functions

Goals / “I can…”I can graph quadratic functions of the form

y = ax + bx + cI can graph quadratic inequalities

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Yesterday we learned about y = axy = ax and y = ax + cy = ax + c. The aa changes the ????? and the cc moves the parabola ?????.

2

2


Today we look at y = ax + bx + c. The bx moves the parabola horizontally, changing the location of the line of symmetry.

2


In an equation y = ax + bx + c, the line of symmetry can be found by the equation

The x-coordinate of the vertex is

2

a

bx

2

a

b

2

Find the line of symmetry of y = 3x2 – 18x + 7

Finding the Line of SymmetryWhen a quadratic function is in standard form

The equation of the line of symmetry is

y = ax2 + bx + c,

2ba

x

For example…

Using the formula…

This is best read as …

the opposite of b divided by the quantity of 2 times a.

18

2 3x 18

6 3

Thus, the line of symmetry is xx = 3 = 3


Finding the VertexWe know the line of symmetry always goes through the vertex.

Thus, the line of symmetry gives us the x – coordinate of the vertex.

To find the y – coordinate of the vertex, we need to plug the x – value into the original equation.

STEP 1: Find the line of symmetry

STEP 2: Plug the x – value into the original equation to find the y value.

y = –2x2 + 8x –3

8 8 22 2( 2) 4

ba

x

y = –2(2)2 + 8(2) –3

y = –2(4)+ 8(2) –3

y = –8+ 16 –3

y = 5

Therefore, the vertex is (2 , 5)


A Quadratic Function in Standard FormThe standard form of a quadratic function is given by

yy = = axax22 + + bxbx + + cc

There are 3 steps to graphing a parabola in standard form.

STEP 1: Find the line of symmetry

STEP 2: Find the vertex

STEP 3: Find two other points and reflect them across the line of symmetry. Then connect the five points with a smooth curve.

Plug in the line of symmetry (x – value) to

obtain the y – value of the vertex.

MAKE A TABLE

using x – values close to the line of symmetry.

USE the equation

2bxa

-=



Graph the equation

y = -3x + 6x + 52


Graphing Quadratic InequalitiesGraphing a parabola with an inequalityinequality is

similar to a line. We use solid and dashed lines and we shade above (greater) or below (less) the line.

Graphs will look like a parabola with a solid or dotted line and a shaded section.

The graph could be shaded inside the parabola or outside.

Forms of Quadratic InequalitiesForms of Quadratic Inequalitiesy<ax2+bx+c y>ax2+bx+cy≤ax2+bx+c y≥ax2+bx+c

1. Sketch the parabola y=ax2+bx+c

(dotted line for < or >, solid line for ≤ or ≥)

** remember to use 5 points for the graph!

2. Choose a test point and see whether it is a solution of the inequality.

3. Shade the appropriate region.

(if the point is a solution, shade where the point is, if it’s not a solution, shade the other region)


Graph y ≤ x2 + 6x - 4

3)1(2

6

2

a

bx

* Vertex: (-3,-13)

* Opens up, solid line

134189

4)3(6)3( 2

y 9- 5-

12- 4-

13- 3-

12- 2-

9- 1-

yx

•Test Point: (0,0)

0≤02+6(0)-4

0≤-4 So, shade where the point is NOT!

Test point


Graph: y > -x2 + 4x - 3* Opens down, dotted

line.

* Vertex: (2,1)

2)1(2

4

2

a

bx

1384

3)2(4)2(1 2

y

y

* Test point (0,0)

0>-02+4(0)-3

0>-3

x y

0 -3

1 0

2 1

3 0

4 -3

Test Point


1) y ≥ x2 and 2) y ≤ -x2 + 2x + 4 Graph bothboth on the same

coordinate plane. The place where the shadings overlap is the solution.

Vertex of #1: (0,0)Other points: (-2,4), (-1,1), (1,1),

(2,4)

Vertex of #2: (1,5)Other points: (-1,1), (0,4), (2,4),

(3,1)

* Test point (1,0): doesn’t work in #1, works in #2.

SOLUTION!



Graph:

y > x + x+ 12

10.2 Quadratic Functions. 10.2 – Quadratic Functions Goals / “I can…” I can graph quadratic...

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