Quadratic EquationsQuadratic Equations

Unit 2Unit 2

English CasbarroEnglish Casbarro

Section 1-Section 1- Graphing Quadratic EquationsGraphing Quadratic EquationsQuadratic Equations are equations in the form Quadratic Equations are equations in the form

The graphs of quadratic equations are called parabolas.The graphs of quadratic equations are called parabolas.

0,2 acbxaxy

If a is positive, the parabola points up, ; if a is negative, the parabola points down,

The following graphs show the vertex (–3, 0) of one parabola (which is the only solution), and the solutions (1, 0) and (4, 0) of another parabola. The third parabola has imaginary solutions.

Graphing from Standard FormGraphing from Standard Form

Standard and Vertex form follow the sameStandard and Vertex form follow the same

rules. There are 3 things you always find:rules. There are 3 things you always find:

The vertexThe vertexThe y-interceptThe y-interceptThe matching pointThe matching point

1. Finding the vertex:

You must use the formula: ,so that you can find the axis of symmetry and the x-value of the vertex.

Next, you substitute the x into the equation to find the y-value of the vertex.

So the vertex is (1,5)

Standard Form:Standard Form: 742 2 xxy

ab

x2

144

)2(2)4(

x

57427)1(4)1(2 2 y

From the previous slide, we found the vertex of the parabola, using the formula:

and substituting the x value into the equation to find the y-value of the vertex. You found that the vertex is

(1,5).

The “a” from your equation is the same “a” in the “a” in standard form. So,

The equation in vertex form is:

Vertex Form:Vertex Form: khxay 2)(

ab

x2

5)1(2 2 xy

Vertex form without completing the square

Now that you can find the vertex of any quadratic function, it is easy to put the function in vertex form.

Ex. Put in vertex form.

Find So, Now,

So, the vertex is (1,8) and a = 3.

The equation in vertex form is:

1163 2 xxy

a

bx

2

1

6

6

)3(2

)6(

x

8113116311)1(6)1(3 2

8)1(3 2 xy

Section 7: Quadratic InequalitiesSection 7: Quadratic Inequalities

You have graphed in all 3 forms of Quadratic You have graphed in all 3 forms of Quadratic Equations:Equations:

Standard FormStandard Form Vertex FormVertex Form Solutions (factored) FormSolutions (factored) Form

Now you will graph the parabolas, then shade Now you will graph the parabolas, then shade for the inequalities.for the inequalities.

GraphingGraphing

You must still find the 3 points to graph:You must still find the 3 points to graph:

The vertexThe vertexThe y-interceptThe y-interceptThe matching pointThe matching point

Example 1: 32 2 xxy

First, we find the vertex:

81

2

3825

341

161

2341

41

2

41

)2(2)1(

2

y

ab

x

Next, we find the y-intercept:Next, we find the y-intercept:

Substitute 0 into the inequality:Substitute 0 into the inequality:

The y-intercept is (0,3).The y-intercept is (0,3).

3

30)0(2 2

y

y

The matching pointThe matching point

The vertex is:The vertex is:

The y-intercept is: The y-intercept is:

The matching point: The matching point:

Graph these 3 pointsGraph these 3 points

825,

41

3,0

3,

21

ShadingShading

The inequality was:The inequality was:

We always shade “above” the graph We always shade “above” the graph forfor

The equal sign below it means that the The equal sign below it means that the line is solid. line is solid.

32 2 xxy

GraphGraph

Using the calculatorUsing the calculator

Go to Y=Go to Y= Use the direction buttons to move as far to the left Use the direction buttons to move as far to the left

as you canas you can Use the “Enter” button to change the type of Use the “Enter” button to change the type of

displaydisplay is represented by the symbol is represented by the symbol Type in the quadratic inequalityType in the quadratic inequality Graph the inequality: the shading will be Graph the inequality: the shading will be

automatic.automatic.

Section 8: ModelingSection 8: Modeling

The vertex form of a quadratic equation is:The vertex form of a quadratic equation is:y + a(x – h)y + a(x – h)22 + k + k

You can use this form to find equations if You can use this form to find equations if you are given pointsyou are given points

Given the vertex and a pointGiven the vertex and a point

Ex. 1: (#10 in your book)Ex. 1: (#10 in your book)Vertex: (2, –1)Vertex: (2, –1)Point: (4, 3) Point: (4, 3)

The equation is y = a(x – h)The equation is y = a(x – h)22 + k + k This means that you are given, as the question, 4 This means that you are given, as the question, 4

out of the 5 variables in the general equation. out of the 5 variables in the general equation. (h = 2, k = –1)(h = 2, k = –1)(x = 4, y = 3)(x = 4, y = 3)

Fill in the general equationFill in the general equation

y = a(x – h)y = a(x – h)22 + k + k

3 = a(4 – 2)3 = a(4 – 2)22 + (–1) + (–1) (substitute h, k, x, and (substitute h, k, x, and y)y)

3 = a(2)3 = a(2)22 – 1 – 1

3 = 4a – 13 = 4a – 1 (add 1 to both sides) (add 1 to both sides)

4 = 4a4 = 4a (divide both sides by 4) (divide both sides by 4)

a = 1a = 1

Fill in the equation with h, k, aFill in the equation with h, k, a

The general equation is : y = a(x – h)The general equation is : y = a(x – h)22 + k + k

(h, k) (h, k) (2, –1) (2, –1)

a = 1 a = 1

So the specific equation is:So the specific equation is:

y = 1(x – 2)y = 1(x – 2)22 – 1 – 1

You try this: You try this:

Vertex: (4, 5)Vertex: (4, 5)

Point: (8, –3) Point: (8, –3)

Solution:Solution:

y = a(x – h)y = a(x – h)22 + k + k Substitute h, k, x, and ySubstitute h, k, x, and y

––3 = a(8 – 4)3 = a(8 – 4)22 + 5 + 5

––3 = a(4)3 = a(4)2 2 +5 +5 Subtract 5 from both sides Subtract 5 from both sides

––8 = 16a8 = 16a Divide both sides by 16Divide both sides by 16

a = -1/2 a = -1/2

So the specific equation is y = –1/2(x – 4)So the specific equation is y = –1/2(x – 4)22 + 5 + 5