# Quadratic Functions

by
**aarthur**

Last updated 8 years ago

** Discipline: **

Math ** Subject: **

Algebra I
** Grade:**

8

Quadratic Poster Project

Quadratic Function # 1x^2+6x+5This will be the minimum vertex point equation.

X Y -6 5 -5 0 -4 -3 -3 -4 -2 -3 -1 0 0 5

To find the axis of symmetry by hand, you use the formula x= -b/2a for the equation ax^2+bx+c. The final formula would be -(6)/2(1). This is equal to -3. To find the vertex by hand, you put the AoS in the quadratic function, and solve for Y. (-3)^2+6(-3)+5. The next step would be to multiply and get 9-18+5. After that you add like terms and get -4 as your y value answer. This makes the vertex (-3,-4).

Quadratic Function # 2-x^2+4x-1This will be the function with the maximum vertex point.

X Y-1 -6 0 -1 1 2 2 3 3 2 4 -1 5 -6

To find if a graph opens either up or down using the equation you see if the a value is positive or negative. If it is positive in will be minimum (U shaped) and if it is negative it will be maximum(umbrella shaped).

To find the axis of symmetry by hand, you use the formula x= -b/2a for the equation ax^2+bx+c. The final formula would be -(4)/-2(1). This is equal to 2. To find the vertex by hand, you put the AoS in the quadratic function, and solve for Y. -(2)^2+4(2)-1. The next step would be to multiply and get -4+8-1. After that you add like terms and get 3 as your y value answer. This makes the vertex (2,3).

-Axis of Symmetry: -3-Vertex: (-3,-4)-Range: y >= -4

-Domain: all real numbers-Y intercept: 5-Solutions: (-1,0) (-5,0)

-Axis of Symmetry: 2-Vertex: (2,3)-Range: y <= 3

-Domain: all real numbers-Y intercept: -1-Solutions: (.3,0) (3.7,0)

To solve the quadratic equation by factoring the X intercepts you first take your equation( I'll use x^2+10x+16) first create your MA table. Something that would multiply to get 16 and add to get 10 is 8 and 2. You then make 2 binomials out of this and get (x+2) and (x+8). You solve these for 0 and you get -2 and -8 as your x intercepts.

One real life example of a parabola could be a fish jumping out of the water and then landing back in it. Where it jumped out and where it landed back in the water would be the x intercepts. The vertex would be the maximum point the fish reached and the parabola would continue after the fish hit the water because it would keep heading underwater in that direction.

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