# Quadratic Functions

by
**aarthur**

Last updated 7 years ago

** Discipline: **

Math ** Subject: **

Algebra I
** Grade:**

7

Quadratics Project

Real Life Example

A real life example would be someone kicking a soccer ball. The y-axis would be the height the ball went and the x-axis would be the distance. The x-intercepts would be where the ball first started to where it hit the ground.

Real Life Example

Quadratic Function #1:y = 3x^2-2x-5Minimum parabola.

X

Y

-2

-1

1

0

2

11

0

-5

-4

3

Axis of Symmetry: .33Vertex: (.33, -5.3)Range: y >= -5.3

Domain: All Real NumbersSolutions: (-1, 1.8)Y-Intercept: -5

If you want to find the axis of symmetry by hand, you have to use the equation x=(-b/2a). For 3x^2-2x-5 the b value would be -2 and the a value would be 3. So -(-2)/2(3) , or 2/6 would be 1/3, or .33 . For the vertex, you have to use the axis of symmetry as the x value and the solved function as the y value. For this function, the axis of symmetry is .33 . If you replace .33 in the function, (3(.33)^2-2(.33)-5) you would get -5.3 as the final value. This means the vertex is (.33, -5.3).

Quadratic Function #2:y = -8x^2+10x+3Maximum Parabola.

Axis of Symmetry: .625Vertex: (.625, 6.125)Range: y <= 6.125

Domain: All Real NumbersSolutions: (-.4, 1.4)Y-Intercept: 3

If you want to find the axis of symmetry by hand, you have to use the equation x=(-b/2a). For -8x^2+10x+3 the b value would be 10 and the a value would be -8. So -(10)/2(-8) , or -10/-16 would be 5/8, or .625 . For the vertex, you have to use the axis of symmetry as the x value and the solved function as the y value. For this function, the axis of symmetry is .625 . If you replace .625 in the function, (-8(.625)^2+10(.625)+3) you would get 6.125 as the final value. This means the vertex is (.625, 6.125).

X

Y

-2

-1

0

1

2

-49

-15

3

5

-9

Parabolas have either a minimum or maximum point and direction. To tell if a graph opens up or down from the function you need to look at the a value. In each quadratic function with a negative a value, the parabola is always maximum. If the a value is positive, it is a minimum parabola.

Minimum/Maximum

Quadratic Equation:y = ax^2+bx+c

To find the solutions, or x-intercepts, you must factor the function. For example, I'll use the quadratic function y = 3x^2-2x-5. The first step is to find two numbers that add up to the b value and multiply to get the product of the a and c value. In this case, the a value is 3, the b value is -2, and the c value is -5. This means we have to find two numbers that add up to -2 and multiply to get -15. This would be 5 and -3. Next we rewrite our function and replace the b value with the two numbers. 3x^2+5x-3x-5. Then we factor it. x(3x+5)-1(3x+5) -> (x-1)(3x+5). Then we take that final form and solve it. The x-intercepts would be -1 and 1.8.

Solutions/X-Intercepts

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