Quadratic Functions

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by aarthur
Last updated 5 years ago

Discipline:
Math
Subject:
Algebra I
Grade:
7

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Quadratic Functions

Quadratic Funchtion 1

Vertex) (0,0)AOS)X=0Soulutions)(0,0)Y intercept)(0,0)Domain) All real #sRange)Y>0

Y=2x^2

To find the Axis of Symmetry(AOS) you must find -b/2a. In a quadratic equation like this one you will use -b as 0 and 2a as 4. You will then get 0/4 which equals 0. 0 will be the AOS line on the y axis.To find the vertex of this equation you take the -b/2a answer and put it into your equation for x. So your equaion will be set up like this: 2(0)^2. When you put toghether tid equation you will find that the answer to the equation once you put 0 in for x is 0 again. Then you take the -b/2a answer and pu that as your x in the coorinate and put the simplified equation in for the y in the coordinate. After doing this you will come up with your vertex. In this equation you will find that your vertex is (0,0).To find the X intercepts or solutions you will factor this equation and set it to zero. First you take the c term in the equation and multiply it by the a term. You will get 0x2 which equals 0. then you take your b term and your CxA number and create a MA table. The only thing that mulitplies to 0 and adds up to 0 is, well, 0 and 0. Then you must put your 0s in for your b term. Your equaiton will be set up like this:2x^2+0x+0x+0. Then you set up each sids like this (2x^2+0x)(0x+0). After symplifying this you should come out with your x intercept(s) being (0,0).

(x,y)(-6,72)(-3,18)(0,0)(2,8)(4,32)

You can tell that the graph opens up because in the original equation the A term is positive. You can also tell it opens up because none of the y terms in the table are negative.

Quadratic Funtion 2

Y=-x^2+16

Vertex)(0,16)AOS)X=0Slutions)(0,-4) and (0,4)Y intercept)(0,16)Domain)all real #sRange Y<16

A Rainbow is a good example of a real life downward opening porabola. The rainbow increases from the left to the middle and decreses from the middle to the left.

(x,y)(-3,-20)(-4,0)(-2,12)(0,16)(4,0)(3,-20)

You can tell that this prabola will open down because in the original equation the a term in negative. Also you can tell because the coordinates in the table go from negative to positive and back to negative.

Deer antlers are a good example of a real life upward opening porabola. the antlers come down from the tips of the points on the antlers to just above the middle of the head.

To find the AOS of the equation -x^2+16 you must first find your -b/2a. In this equation that would be -0/-2. That would equal 0 this is the line on the y axis at which the AOS falls on.To find the vertex of this equation you need to plug i your -b/2a back into your eqution. After doing this you will set your equation up like this:-0^2+16. This equals 16. Then you put your -b/2a (0) in for your x in your coordinate and your 16 in for your y. Your coordinate will look like this (0,16) and that will be your vertex.You can find the X intercepts by solving the equation. First you take you c term multiplied by your a term. after this you will get -16. You will then need to set uo a MA table using your b term and your AxC answer. the two number that you will get that add up to 0 and mulitiply to -16 are 4 and -4. You will need to put these numbers into your equation for your b term. Your equaiton will be set up like this: -x^2-4x+4x+16. Then you seperate th eequation like this: (-x^2-4x)(4x+16). After factoring these two monomials you should come up with your X intercepts as (-4,0) and (4,0)


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