Exercise 171 Page 115

a Examining the function, we notice that it's an absolute value function. The general equation of such a function is written in the following format.

General Equation:& y=a|x-h|+k Locator Point:& (h,k) If we rewrite our equation to match this form exactly, we can identify the locator point.

Equation:& f(x)=- |x-(-2)|+(-1) Locator Point:& (-2,- 1) To graph the function, we need a few more data points that fall on the graph. |c|r|c| [-1em] x & -|x+2|-1 & f(x) [0.2em] [-1em] -6 & -| -6+2| - 1 & -5 [0.2em] [-1em] -4 & -| -4+2| - 1 & -3 [0.2em] [-1em] 0 & -| 0+2| - 1 & -3 [0.2em] [-1em] 2 & -| 2+2| - 1 & -5 [0.2em] Let's mark the points in a coordinate plane and draw the graph.

b This equation reflects a hyperbola. The general equation of a hyperbola allows us to identify its locator point, as well as the horizontal and vertical asymptote.

General Equation:& y=a(1/x-h)+k [0.8em] Locator Point:& (h,k) [0.8em] Horizontal Asymptote:& y=k [0.8em] Vertical Asymptote:& x=h If we rewrite our equation to match this format exactly, we can determine the locator point, and asymptotes.

Function:& y=1/x-+2 [0.8em] Locator Point:& ( ,2) [0.8em] Horizontal Asymptote:& y=2 [0.8em] Vertical Asymptote:& x= Let's now draw the graph. To help us do that, we first make a table of values. |c|c|c| [-1em] x & 1/x+2 & y [0.6em] [-1em] -4 & 1/-4+2 & 1.75 [0.6em] [-1em] -2 & 1/-2+2 & 1.5 [0.6em] [-1em] -0.5 & 1/-0.5+2 & 0 [0.6em] [-1em] -0.25 & 1/-0.25+2 & - 2 [0.6em] [-1em] 0.25 & 1/0.25+2 & 6 [0.6em] [-1em] 0.5 & 1/0.5+2 & 5 [0.6em] [-1em] 2 & 1/2+2 & 2.5 [0.6em] [-1em] 4 & 1/4+2 & 2.25 [0.6em] Let's mark these points in a coordinate plane and draw the graph.

c Like in Part B, we will first find the locator point and asymptotes by rewriting the equation so that it exactly matches the general equation of a hyperbola.

Function:& y=1/x-(-5)+(-2) [0.8em] Locator Point:& (-5,-2) [0.8em] Horizontal Asymptote:& y=-2 [0.8em] Vertical Asymptote:& x=-5

Let's now draw the hyperbola's graph. To help us do that, we calculate a few points that fall on the graph in a table of values. |c|c|c| [-1em] x & 1/x+5-2 & y [0.6em] [-1em] -10 & 1/-10+5-2 & -2.2 [0.6em] [-1em] -7 & 1/-7+5-2 & -2.5 [0.6em] [-1em] -5.5 & 1/-5.5+5-2 & -4 [0.6em] [-1em] -5.25 & 1/-5.25+5-2 & - 6 [0.6em] [-1em] -4.75 & 1/-4.75+5-2 & 2 [0.6em] [-1em] -4.5 & 1/-4.5+5-2 & 0 [0.6em] [-1em] -3 & 1/-3+5-2 & -1.5 [0.6em] [-1em] 0 & 1/0+5-2 & -1.8 [0.6em] Let's mark these points in a coordinate plane and draw the graph.

d This is a cubic function. These functions can be described with the following general equation.

General Equation:& y=a(x-h)^3+k Locator Point:& (h,k) If we rewrite our function so that it matches this format exactly, we can determine the locator point.

Function:& y=-(x- )^3+5 Locator Point:& ( ,5) Let's now draw the cubic function's graph. To help us do that, we must calculate a few points that fall on the graph using a table of values. |c|c|c| [-1em] x & - x^3+5 & y [0.2em] [-1em] -2 & - ( - 2)^3+5 & 13 [0.2em] [-1em] -1 & - ( - 1)^3+5 & 6 [0.2em] [-1em] 0 & -( 0)^3+5 & 5 [0.2em] [-1em] 1 & -( 1)^3+5 & 4 [0.2em] [-1em] 2 & -( 2)^3+5 & -3 [0.2em] Next we will mark these points in a coordinate plane and draw the graph.

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Exercise 171 Page 115

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