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Precalc – 2.1 - Quadratics. What do we already know?. Representations. ALGEBRA. POLYNOMIAL FORM. STANDARD FORM. FACTORED FORM. y = ax 2 + bx + c. y = a(x - h) 2 + k. y = ( x + a)(x + b ) . ORDERED PAIRS. GRAPH. Parabola Graph Vocabulary. - PowerPoint PPT Presentation
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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 −
b2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟
⎛ ⎝ ⎜
⎞ ⎠ ⎟
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