Precalc – Quadratics

What do we already know?

Representations ALGEBRA ORDERED PAIRSGRAPH POLYNOMIAL FORMFACTORED FORMSTANDARD FORM y = ax 2 + bx + cy = a(x - h) 2 + ky = (x + a)(x + b)

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) Vertex of the equation y = a(x – h) 2 + k

Practice What are the vertices of these equations? – y = 3(x – 5) – y = 4(x + 9) – y = (x – 7) 2 – y = x – y = x 2 – 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 FORMFACTORED FORMSTANDARD FORM y = ax 2 + bx + cy = a(x - h) 2 + ky = (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(x 2 – 2x + 1) – 8 y = 2x 2 – 4x + 2 – 8 y = 2x 2 – 4x – 6

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

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

How to factor: Given: f(x)=ax 2 +bx+c Find two numbers that: – Add up to b – Multiply out to a x c 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 = 2x 2 – 4x – 6 – y = 2x 2 +3x-5 – y = 6x 2 +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 = 0x – 3 = 0

Practice Graph these functions (vertex from vertex form, zeros from factored form) – y = 2(x – 1) 2 – 8 – y = -(x + 4) – 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 = 2x 2 – 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 = ax 2 + bx + c, the vertex will be: This comes from calculus

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

ALGEBRA GRAPH POLYNOMIAL FORMFACTORED FORMSTANDARD FORM y = ax 2 + bx + cy = a(x - h) 2 + ky = (x + a)(x + b) 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