6.4a Converting Standard and Vertex Forms.notebook
May 05, 2014
Warm Ups(6.4a)
For each equation below:
i) State the coordinates of the vertex
ii) State the equation of the axis of symmetry
iii) State the y­intercept
iv) State which way the parabola opens
a) y = ­ 2x2 ­ 7 b) y = 2.5(x ­ 1)2 ­ 3 c)y = ­ 4(x + 1)2 + 4
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6.4a Converting Standard and Vertex Forms.notebook
May 05, 2014
6.4a Standard and Vertex Forms
Recall a) Standard Form _________________ tells us the _________from___
b) Vertex Form _________________ tells us the _________ from___
We switch forms to help us find out different components of the parabola. To convert Vertex form to Standard form:
a) square the binomial
b) multiply "a" using distributive property
c) collect like terms
Ex. 1. Identify the y­intercept of y = ­3(x ­ 1)2 + 4.
Solution A {sub. x = 0}
Solution B {Expand to standard form}
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6.4a Converting Standard and Vertex Forms.notebook
May 05, 2014
To convert Standard form to Vertex form, there are 3 methods
Method 1. Complete the Square a) common factor "a" out of first two terms
b) In margin....b÷2 and (b÷2)2 c) complete the square (Add and subtract the same value inside the bracket.)
d) collect like terms
In an example it will look like this :
Convert y = 3x2 + 6x + 7
a) common factor "a" out of first two terms
y = 3(x2 + 2x) + 7
b) In margin....b÷2 and (b÷2)2 {(2÷2) = 1} , {12 = 1}
c) complete the square
y = 3(x2 + 2x +1 ­ 1) + 7
y = 3(x2 + 2x + 1) ­ 3 + 7
d) collect like terms
y = 3(x + 1)2 + 4
{­1 x 3 = ­3} (Affect of "a")
Since the vertex is ______ and "a" is ____, this tells us that we have a ________value of _________ when x = _____.
We will learn method 2 and 3 next day!
Ex. 2. Without sketching the graph, determine the maximum or minimum value of the function y = ­2x2 ­ 20x ­ 47. .
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6.4a Converting Standard and Vertex Forms.notebook
May 05, 2014
How can we determine the equation of the parabola in vertex form if we are given the vertex and a point?
Example: Determine the equation of the parabola. a) vertex (0,2), passing through (­3, 11)
Step 1: y = a(x ­ p)2 + q is the equation so we will substitute in our values for p and q which are given to us by the coordinates of the vertex.
y = a(x ­ 0)2 + 2
Step 2: We are given a point so we have an x and a y value. For this example x = ­3 and y = 11
11 = a(­3 ­ 0)2 + 2
Step 3: Solve for a.
11 = a (­3)2 + 2
Step 4: State the general equation with values substituted in for a, p, and q. a =
p =
q =
y = a(x ­ p)2 + q
Now try it on your own! Determine the equation of a parabola whose vertex is (0, 2) and passes through point (2, 9)
HW. Pg. 314 # 1 ­3, 7ace, 8ace, 10ac, 13af
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6.4a Converting Standard and Vertex Forms.notebook
May 05, 2014
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