Graphing+Continued

__**Graphing Continued**__ (and some review stuff)


 * October 11/ 11**

__//Example 1.//__ //Write the Equation represented by the following graph://



Since the graph intersects the x-axis at points (-3, 0), (1,0) and (6,0), these points are what the x value becomes when y=0. Hence, (x+3), (x-1) and (x-6) are factors, and values at which y=0. Judging from the graph, we know that it is a cubic equation as there are no other turning points or irregularities within the curves. Therefore all factors of 'x' have been found. From this we can formulate:

y=a(x+3)(x-1)(x-6)

Since there is a non-zero 'y' value present, it can be plugged into the formula, rendering:

-6=a(x+3)(x-1)(x-6) a= -1/3 Therefore: y=-1/3(x+3)(x-1)(x-6)

__//Example 2//__

//Write the Equation represented by the following graph://



//Therefore (x+2) and (x-2) are factors and perfect squares.//

y=a ((x+2)^2) ((x-2)^2) 4=a ((x+2)^2) ((x-2)^2) a=1/4

Therefore, y=1/4 ((x+2)^2) ((x-2)^2)