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**Nonlinear Functions**

y = mx + b 3x + 5y - 10 = 0 y = 88x are all examples of linear equations.The graphs of

In this topic, we will be working with nonlinear functions with the form *y = ax*^{2} + *b* and *y = ax*^{3} *b* where *a* and *b* are integers.

* Quadratic functions: y = ax ^{2} + b*

The graph of the functiony = ax^{2}+bwill look like a "U". This "U" shape graph is called aparabola. Whenais positive, then the parabola opens up. Whenais negative, then the parabola opens down.The highest or lowest point of parabolas is called the

vertex.bdetermines where the vertex is on the graph. Whenb=0, the vertex is on the origin (0,0). Whenb = hwherehis an integer, the vertex is on the point (0,h).

In this graph, the vertex is the lowest point.

b = 0 because the vertex is on the origin.Use the point (2,12) to find

a.

y = ax ^{2}12 = a(2) ^{2}12 = 4a 3 = a The equation is y = 3x.

In this graph, the vertex is the highest point.

b = -5 because the vertex is on (0, -5).Use the point (1, -7) to find

a.

y = ax ^{2}- 5-7 = a(1) ^{2}- 5-7 = a - 5 -2 = a The equation is y = -2x - 5.

* Cubic functions: y = ax ^{3} + b*

The graph of a cubic function has this shape

b = 0 when the point of transition (from an upwards curve to a downwards curve) is on the origin (0,0).

This is an example of y = ax^{3}whereais negative.Use the point (1, -2) to find

a.

y = ax ^{3}-2 = a(1) ^{3}-2 = a The equation is y = -2x

^{3}.

b = -5 because the point of transition is on (0, -5). Use the point (1, -2) to find

a.

y = ax ^{3}- 5-2 = a(1) ^{3}- 5-2 = a - 5 3 = a The equation is y = 3x

^{3}- 5.

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