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# 06.05 Graphing Systems of Nonlinear Equations

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## Mina Farid

on 12 December 2015

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#### Transcript of 06.05 Graphing Systems of Nonlinear Equations

06.05 Graphing Systems of Nonlinear Equations

Create a table of values for a linear function.
The domain of f(x) is (-∞, ∞)
The range of f(x) is (-∞, 36)

These are important because they tell you where your graph lies and whether or not it continues on forever on the x and y axis. all values of x don't make sense for the application. For example, if you were to consider the x-axis as the ground (or whatever surface the rainbow starts and ends on), then the domain for this application is [-6, 6]. Similarly the range of f(x) is (-∞, 36), but only [0, 36] make sense.

X-INT: (6,0) , (-6,0)
Y-INT: (0,36)

Its positive because it goes up into the "rainbow"

PART 3
Now for the second part, just pick any two points with which we can draw a line with a positive slope. I'll use x = -2 and 1:

y = -({-2})^2 + 36 = {-4} + 36 = 32
y = -(1)^2 + 36 = {-1} + 36 = 35

So, our two points are (-2, 32) and (1, 35). To find the equation of the linear function that goes through these two points, let's use slope-intercept form, which is f(x) = mx + b. The slope m is given by \frac{y_2 - y_1}{x_2 - x_1}, so

m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{35 - 32}{1 - ({-2})} = 1
So, the equation of the linear function so far is just f(x) = x + b, and we can find b by plugging in one of the points on the line:

35 = 1 + b
b = 34

Thus, the equation of the linear function is f(x) = x + 34

x I y
----------
-2 32
-1 35
0 36
1 35
2 32
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