5-Chris Rowland

by aarthur
Last updated 7 years ago

Algebra I

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5-Chris Rowland

Porabolas Project

The slinky above resembles a maximum vertex porabola because of its unique u shape. The x intercepts would be where the slinky rests on the table. The vertex wolud be the maximum point the slinky reaches in its arch.

To tell if a graph opens up or down you will need to look at the "a" value. If the "a" value is positive the graph opens upwards. If the "a" value is negative the graph opens downwards.

To find the AoS you will need to follow the equation (-)b/(2)a. For example to find the AoS for the equation 2x^2+4x+6 you would take the "a" value (2) and the "b" value (4) and plug them into the equation (-)b/(2)a. Once you get your equation -4/4 you solve for your answer. The AoS for this example is -1.AoS = -1

To find the vertex of an equation you will first need to find the AoS. For example the AoS of the equation 2x^2+4x+6 is -1. Once you have found your AoS you will substitute it for the unknown (x) value in your equation. -4-4+6. Once you have imputed the AoS into the equation you solve for your answer, in this case the answer is -2. So our vertex is going to be the AoS and the answer you receive after solving the equation. Vertex = (-1,-2)

Quadratic Function #1 -3x^2+4This will be the the maximum quadratic function.

AoS: 0Vertex: (0,4)Solutions: (0,-1.32), (0,1.47)Y Intercept: (0,4)Domain: All real numbersRange: x > 4

Quadratic Function #2 x^2-8x+5This will be the minimum quadratic function

AoS: 4Vertex: (4,-11)Solutions: (0,-2.6), (0,11.69)Y Intercept: (0,5)Domain: All real numbersRange: x > -11



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