4.2 Quadratic Functions β Vertex and Intercept Form...2+ + 1. Opens up if >0,Opens down if
Transcript of 4.2 Quadratic Functions β Vertex and Intercept Form...2+ + 1. Opens up if >0,Opens down if
4.2 Quadratic Functions βVertex and Intercept Form
ππ₯2 + ππ₯ + π1. Opens up if π > 0, Opens down if π < 0
2. The axis of symmetry is π = βπ
ππ
3. The vertex has the x-coordinate βπ
ππ: β
π
ππ, π(β
π
ππ)
4. The y-intercept is π, π, π
5. How might we determine the Max and Min of a quadratic equation?
WARM UP! Hot Potato with your partner.
Graph the following functions
π = ππ β ππ + π π π = ππ β ππ + π
π = π β π β π π π = βπ π + π + π
Graph the following functions
Graph the following functions
π = ππ β ππ + ππ π = ππ β ππ + ππ = π β π β ππ π = βπ π + π + π
Vertex Form π = π(π β π)π+π
Using what you already know, how might you graph the function -
π = βπ(π β π)π+π
Vertex Form π¦ = π(π₯ β β)2+π
Parent Equation π¦ = ππ₯2π¦ = (π₯ β 2)2+1π¦ = (π β π)2+1π¦ = (π β π)2+π
Vertex (π , π)
Steps to graph from vertex form
1. Identify the constants π = π(π β π)π+π
2. Does it open up or down? Is π < 0 ππ π > 0
3. Plot the vertex (π , π)
4. Evaluate a two points symmetric about π
5. Plot the graph
You Try!
π¦ = (π₯ + 4)2 π¦ = 2(π₯ + 1)2 β 3
Intercept Form π¦ = π(π₯ β π)(π₯ β π)
β’ x-intercepts are π πππ π
β’ π πππ π are symmetric about the vertex
π₯ =π+π
2
β’ The vertex is atπ+π
π, π
π+π
2, π
π+π
2
π¦ =1
2(π₯ β 1)(π₯ β 5)π¦ =
1
2(π₯ β 1)(π₯ β 5)
Steps to graph from intercept form
1. Does it open up or down? Is π < 0 ππ π > 0
2. Identify the x-intercepts (π , 0) and π , 0
3. Axis of Symmetry is π₯ =π+π
2
4. Plot the vertex π+π
2, f(
π+π
2)
5. Plot the graph
You Try!
π¦ = 2(π₯ β 1)(π₯ β 4) π¦ = β2(π₯ + 2)(π₯ β 4)
Use the FOIL method to multiply the binomials
First
Outside
Inside
Last
+ππ +ππ+ππ+ππ
ππ + ππ + ππ
(π + π)(π + π)
π¦ = 2(π₯ β 2)(π₯ + 1) π¦ = 2(π₯ β 2)2+2
Converting to Standard Form
Homework
Section 4.2 1 β 53 odd