By Dawkins P.

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**Extra resources for Algebra/Trig Review (2006)(en)(98s)**

**Sample text**

To find the solution to this inequality we need to recall that polynomials are nice smooth functions that have no breaks in them. This means that as we are moving across the number line (in any direction) if the value of the polynomial changes sign (say from positive to negative) then it MUST go through zero! So, that means that these two numbers ( x = 5 and x = −2 ) are the ONLY places where the polynomial can change sign. The number line is then divided into three regions. In each region if the inequality is satisfied by one point from that region then it is satisfied for ALL points in that region.

Recall that −1 ≤ cos ( x ) ≤ 1 . So, 1 divided by something less than 1 will be greater than 1. Also, 1 secant. R sec (ω x ) ≥ R ±1 = ±1 and so we get the following ranges out of and R sec (ω x ) ≤ − R 8. y = csc ( x ) Solution 1 For this graph we will have to avoid x’s where sine is zero csc x = . So, the sin x graph of cosecant will not exist for x = , −2π , −π , 0, π , 2π , Here is the graph of cosecant. Cosecant will have the same range as secant. R csc (ω x ) ≥ R and R csc (ω x ) ≤ − R 9.

3. Graph the following two curves and determine where they intersect. x = y2 − 4 y − 8 = x 5 y + 28 Solution Below is the graph of the two functions. If you don’t remember how to graph function in the form x = f ( y ) go back to the Graphing and Common Graphs section for quick refresher. There are two intersection points for us to find. In this case, since both equations are of the form x = f ( y ) we’ll just set the two equations equal and solve for y. y 2 − 4 y − 8 = 5 y + 28 y 2 − 9 y − 36 = 0 0 ( y + 3)( y − 12 ) = The y coordinates of the two intersection points are then y = −3 and y = 12 .