Converting to Standard Form

Converting to Standard Form
What’s Standard Form?
Quadratic equations in Standard Form look like this:
𝑦 = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐
Note that π‘Ž β‰  0, and the terms are written from greatest to least degree (i.e. the π‘₯ 2 term is first).
Vertex Form to Standard Form
Quadratics in Vertex Form look like this:
𝑦 = π‘Ž(π‘₯ βˆ’ β„Ž)2 + π‘˜
Warning: Common Mistake!
The binomial (π‘₯ βˆ’ β„Ž) is squared and multiplied by π‘Ž, then π‘˜ is
added.
You can also rewrite this if it helps you to accurately expand it:
Be careful not to multiply by π‘Ž first – you
must square the binomial first!
𝑦 = π‘Ž(π‘₯ βˆ’ β„Ž)(π‘₯ βˆ’ β„Ž) + π‘˜
Example 1
Convert the following quadratic equations from vertex form to standard form:
𝑦 = (π‘₯ βˆ’ 2)2 + 3
𝑦 = 2(π‘₯ + 4)2 βˆ’ 5
When we square a binomial, such as (𝑝 + π‘ž)2 , the result follows a pattern:
(𝑝 + π‘ž)2 = 𝑝2 + 2π‘π‘ž + π‘ž 2
When we’re confident, we can use this pattern to save some time converting.
Example 2
Convert the following quadratic equations from vertex form to standard form using the pattern for squaring a
binomial:
𝑦 = βˆ’3(π‘₯ + 2)2 + 11
𝑦 = 4(π‘₯ βˆ’ 1)2 + 3
Factored Form to Standard Form
Quadratics in Factored Form look like this:
𝑦 = π‘Ž(π‘₯ βˆ’ π‘Ÿ)(π‘₯ βˆ’ 𝑠)
Warning: Common Mistake!
You can either
ο‚·
ο‚·
Be careful not to multiply twice by π‘Ž!
Distribute π‘Ž into the first root only; or
Expand the binomials and then multiply by π‘Ž.
Example 3
Convert the following quadratic equations from factored form to standard form:
𝑦 = (π‘₯ βˆ’ 3)(π‘₯ + 2)
𝑦 = 2(π‘₯ + 4)(π‘₯ βˆ’ 1)
Example 4
Sometimes there is a fractional π‘Ž-value, so it might be best to wait until you’ve expanded the binomials and then
multiply by π‘Ž.
Convert the following quadratic equations from factored form to standard form:
1
𝑦 = 4 (π‘₯ βˆ’ 5)(π‘₯ + 2)
2
𝑦 = βˆ’ 3 (π‘₯ βˆ’ 3)(π‘₯ + 3)
Not-So-Friendly Equations
These equations don’t look as nice, but they work the same way. No need to fear!
Example 5
1 2
2
𝑦 = 4 (π‘₯ + ) +
7
8
𝑦 = 12 βˆ’ 5(π‘₯ + 3)2
𝑦 = (3π‘₯ + 2)(7π‘₯ βˆ’ 0.2)
𝑦 = (5 βˆ’ 2π‘₯)(6π‘₯ + 1)
β„Ž = βˆ’4.9𝑑(𝑑 βˆ’ 7)
𝑃 = 1000 βˆ’ 3(π‘ž βˆ’ 25)2
Summary
No matter how the equation is written, follow the rules of BEDMAS to expand, simplify by collecting like terms, and
write the final polynomial with terms from greatest to least degree.
Practice
Convert each of the following equations into Standard Form.
1
3
a) 𝑦 = (π‘₯ + 2)2 + 2
g) 𝑦 = βˆ’ 2 (π‘₯ βˆ’ 3)2 + 2
b) 𝑦 = 3(π‘₯ βˆ’ 2)2
h) 3π‘₯ 2 βˆ’ 2𝑦 = 0
c) 𝑦 = βˆ’(π‘₯ + 7)2 + 49
i)
𝑦 = 12 βˆ’ (π‘₯ βˆ’ 3)2
d) 𝑦 = 3π‘₯ 2 βˆ’ 17
j)
𝑦=
e) 𝑦 = (π‘₯ + 2)(π‘₯ + 9)
f)
𝑦 = βˆ’3(π‘₯ βˆ’ 2)(π‘₯ + 1)
(π‘₯+4)2
7
βˆ’2
k) 𝑦 = 5(π‘₯ + 5)2 + 5
l)
4
1
𝑦 = 3 (π‘₯ + 3)2 + 6
Make up three quadratic equations in vertex form and three in factored form. Trade with a classmate and convert
each other’s equations into standard form.