# Quadratic Functions

by
**aarthur**

Last updated 7 years ago

** Discipline: **

Math ** Subject: **

Algebra I
** Grade:**

7

y=-3x^2-2X+1Range: Y<1 1/3AOS: -1/3y intercept: (0,1)Solution, x intercept: (.21,0)(-1.06,0)Vertex: (-1/3,1 1/3)Domain: All real numbers

QUADRATIC FUNCTIONS

y= 4x^2-3xRange: y> -9/16AOS: 3/8y intercept: (0,0)Solution, x intercept: (0,0)(.85,0)Vertex: (3/8,-9/16)Domain: All real numbers

On this graugh above you can tell if it opens up or not by the solution: y= 4x^2-3x, all you have to do is see if the x is possitive or negative. If possitive the graugh will have and upward u shape(above) which is opening up, if negative it will have a n shape(left) which is opening down. The graugh above is opening up because the x is possitve.

You can tell if this graugh above opens up or down by the solution: y= -3x^2-2x+8, all you have to do is see if x is positive or negavive. In this solution x is negative which makes the graugh go down.

The way you find axis of symetry is -b/2ay= -3x^2-2x+1-3(x^2)-2x+1-3/9-2/3+12/2(-3) 2/-6 1/-3AOS= 1/-3

The way you find axis of symetry is doing -b/2ay= 4x^2-3x 4(x^2)-3(x).14 - 1 1/8AOS= 3/8

How to find Vertex for 4x^2-3xAOS always comes first(3/8)Then the range always comes secondy(-9/16)Vertex= (3/8,-9/16)

The vertex is the highest or lowest point of the equation.The first number is always the AOS and the second is always your range.

How to find the Vertex for -3x^2-2x+1AOS always comes first(-1/3)Then the rangwe goes after AOS(1 1/3)Vertex= (-1/3, 1 1/3

When solving for the x-intercept you have to factor. The first step is your y-intercept is always the first number(0,1). Secondly you have to put y=0= -3x^2-2x+1(equation). When doing the process you will eventually get (.21,0)(-1.06,0).

When solving for the x-intercept you have to factor. The first step is your y-intercept is always the first numbers(0,0). Secondly you have to put y=0 = 4x^2-3x(ecuation).When doing the process you will eventually get (0,0) (.85,0).

This picture of a wave represents a real world picture of a parobla. This wave would probably be a downward parubla because of the way the wave is crashing down.

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