Precalc – 2.1 - Quadratics

What do we already know?

RepresentationsALGEBRA

ORDERED PAIRS GRAPH

POLYNOMIAL FORM FACTORED FORMSTANDARD FORM

y = ax2 + bx + c y = a(x - h)2 + k y = (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)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 FACTORED FORMSTANDARD 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.

€

x =−b ± b2 − 4ac

2a

Polynomial Graph

• We can also find the vertex• Given the equation y = ax2 + bx + c, the vertex will be:

• This comes from calculus

€

−b

2a, f −

b

2a

⎛

⎝ ⎜

⎞

⎠ ⎟

⎛

⎝ ⎜

⎞

⎠ ⎟

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

GRAPH

POLYNOMIAL FORM FACTORED FORMSTANDARD FORM

y = ax2 + bx + c y = a(x - h)2 + k y = (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