Precalc – 2.1 - Quadratics

Precalc – 2.1 - Quadratics

What do we already know?

Representations ALGEBRA POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) ORDERED PAIRS GRAPH

Parabola Graph Vocabulary • Axis of symmetry: the line that splits the parabola in half • Vertex: the points where the axis of symmetry intersects the parabola; also, either the highest or the lowest point on the graph • Opens upward: graph will look like a U • Opens downward: graph will look like an upside down U

Algebraic  Graphic representation • Moving between algebraic and graphical representations is difficult because different representations give you different types of information • Polynomial form: vertex, opens up or down, zeros • but through the use of formulas • Standard form: vertex, opens up or down, skinny/wide • Factored form: zeros • note: not every parabola has zeros

Standard Form  Graph • You already know how to do this! • Graph: y = -2(x – 2)2 + 4 • Vertex of the equation y = a(x – h)2 + k

Practice • What are the vertices of these equations? • y = 3(x – 5)2 + 9 • y = 4(x + 9)2 + 7 • y = (x – 7)2 • y = x2 + 7 • y = x2 • y = 2(x + 10)2

Standard  Graph • How will we know where its zeros are? • We can only get that information from the polynomial or factored form • *As always, we set y=0, solve for x

Moving between algebraic forms POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) Notice that to move from standard to factored forms, we have to pass through the polynomial form

Standard  Polynomial • Expand the squared section and combine like terms • Ex: y = 2(x – 1)2 – 8 y = 2(x – 1)2 – 8 y = 2(x2 – 2x + 1) – 8 y = 2x2 – 4x + 2 – 8 y = 2x2 – 4x – 6

Practice • Turn these into polynomial form • y = 3(x – 5)2 + 9 • y = 4(x + 9)2 + 7 • y = (x – 7)2 • y = 2(x + 10)2

Polynomial  Factored • Factor! • y = 2x2 – 4x – 6 hm… remember how to factor?

How to factor: • Given: f(x)=ax2+bx+c • Find two numbers that: • Add up to b • Multiply out to axc • Rewrite the equation with those numbers • Pair up your terms and find common factors • Factor the pairs • Find the common factors of the factored pairs • Rewrite as a product of two binomials

Practice • Factor these equations • y = 2x2– 4x – 6 • y = 2x2+3x-5 • y = 6x2+5x+1

Reminder: • Why did we want to change into polynomial and factored forms? • Oh yes – because we can’t get the zeros from standard form • We need the zeros to graph a parabola • Okay! So let’s get those zeros.

Factored  Graph • y = (x + a)(x + b) • The zeros are: • In our example: y = (2x + 2)(x – 3) 2x + 2 = 0 x – 3 = 0

Practice • Graph these functions (vertex from vertex form, zeros from factored form) • y = 2(x – 1)2 – 8 • y = -(x + 4)2 + 9 • y = 4(x – 2)2 - 4

Polynomial  Graph • We didn’t have to factor the equation to get the zeros; there is one other option • 0 = 2x2 – 4x – 6 • Quadratic Formula! Tells the values of x that make the function zero.

Polynomial  Graph • We can also find the vertex • Given the equation y = ax2 + bx + c, the vertex will be: • This comes from calculus

Practice • Graph these functions (zeros from quadratic formula, zeros from –b/2a) • y = 2x2 – 4x – 6 • y = -x2 – 8x – 7 • y = 4x2 – 16x + 12

ALGEBRA POLYNOMIAL FORM STANDARD FORM FACTORED FORM y = ax2 + bx + c y = a(x - h)2 + k y = (x + a)(x + b) GRAPH Put it together: how do we move between all these different representations of quadratic functions?

Exceptions • Why will we have difficulty with a function that has a graph like this?

Graph  Vertex Form

HOMEWORK! • Page 208 #1-8, 13, 18, 20, 23, 24, 26

Precalc – 2.1 - Quadratics