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standard form

y=ax^2+bx+c

factored form

y=a(x-p)(x-q)

vertex form

y=a(x-h)^2+v

vertex= (h,v)

vertex= (h,v)

How can you solve from vertex form?

Completing the square

How can you solve from factored form?

Factoring

How can you solve from standard form?

Quadratic formula*

*when 0=ax^2+bx+c

*when 0=ax^2+bx+c

discriminant=

b^2-4ac (what's under the root)

How to find the x-coordinate of vertex:

from standard: -b/2a

from factored: (p-q)/2

from factored: (p-q)/2

the discriminant is used in order to find...

how many x-ints there are

+ discriminant=2 x-ints

- discriminant= 0 x-ints

when discriminant=0, there is one x-int, which is also the vertex

*b/c you can't sq root a negative number, and root 0 is 0, but root any positive number will have both a positive and negative root

+ discriminant=2 x-ints

- discriminant= 0 x-ints

when discriminant=0, there is one x-int, which is also the vertex

*b/c you can't sq root a negative number, and root 0 is 0, but root any positive number will have both a positive and negative root

How can one identify a quadratic formula?

The second difference: after finding the difference between each term, find the difference between those differences. If that value is a constant number, then we have the second difference, and the formula is quadratic.

We can find a by...

finding the second difference/ 2

2a=second difference

2a=second difference

We know c because...

it's the y intercept (when x is 0)

We can find b by...

once we a & c, we can plug in for x & y with known points, and solve for b