2-3: Solving Quadratic Equations by Factoring Unit 2 English Casbarro.
Quadratic Equations Unit 2 English Casbarro Section 1- Graphing Quadratic Equations Quadratic...
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Transcript of Quadratic Equations Unit 2 English Casbarro Section 1- Graphing Quadratic Equations Quadratic...
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