Algrebra II

hi, thanks for reading this

I have a question for finding vertical and horizontal asymptotes! as to finding the intercepts, domain, and graphing it all together...

like y= 6 over (x+3)

I don't quite understand the concepts and things, but if you can give me a hand that'll be great!



y = 6 / (x+3)
Domain: all possible x values.
Range: all possible y values.
Intercepts: the (x,y) coordinates where the graph intersects the x- and y-axes.
Asymptotes: Horizontal and vertical lines representing x- and y-values that the graph approaches, but never reaches.

I assume that you have a graphing calculator. Set your window so that x and y both range from -10 to +10. Then graph y = 1/x. This is the "inverse" or reciprocal function; for any value x, y is its reciprocal. If x=2, then y=1/2; if x=-7, then y=-1/7. X and y are either both positive or both negative. If the absolute value of x is large, then the absolute value of y is small, and vice versa.

There is one value which neither x nor y will ever be: zero. X can't be zero, because that would be dividing by zero. y=1/x can be manipulated to become xy=1 (multiply both sides by x), and then or x=1/y (from xy=1, divide both sides by y).

Therefore the domain (all possible x values) is: x does not = zero. Ditto the range (all possible y values): y does not = zero.

If you look at the graphs, as a positive x value gets smaller and smaller, it approaches zero, but never reaches it. Similarly, a negative x value can increase and get closer and closer to zero, but never reaches it. x=0 is a value that the graph approaches, but never reaches. We call the line x=0 an asymptote of y=1/x.

Similarly, both positive and negative y values can approach, but never reach 0. Thus, the line y=0 is also an asymptote of y=1/x.

The equation y=mx+b represents a whole "family" of linear equations which have similar properties. Similarly, the y=ax^2+bx+c represents a whole family of quadratic equations.

Similarly, y=1/x is the simplest form of a whole family of "inverse" equations: y=k/(x-h), where k and h are any two numbers except zero, represents a whole family of "inverse" functions.

k "multiplies" the y=1/x equation. Graphically, the y=1/x equation is "vertically stretched". Try k=5. y=5/x. Graph it. You'll see the same 2 curves, but they rise much more steeply near the y axis.

h moves the two curves right by h units. If h=3, then y=1/(x-3). Notice that the denominator has a minus sign, not a plus sign. Since the denominator can't be zero, x-3 can't be zero, so x can't be 3. That means the two curves approach the vertical asymptote x=3. This has the effect of moving both y=1/x curves 3 units to the right. If x = -4, then y=1/(x - (-4)) = 1/(x+4). Now x+4 can't be zero, so x can't = -4, so the vertical asymptote is x=-4.

Can you move the horizontal asymptote up or down? Yes, if you define the basic "family" function as y=k/(x-h) + q. Positive q moves the entire graph upwards, negative q moves it downwards, and the horizontal asmymptote becomes y=q.

For your y=6/(x+3), the domain (x) values are x not = -3, and the vertical asymptote is x = -3. The range (y) values are y not = 0, so the vertical asymptote is y = 0.

How about intercepts? Intercepts are the values where the graph inersects the x- and y-axes. For the y-intercept, x = 0. For your equation, when x=0, y=6/(3+0)=2, so the y-intercept is (0,2). For the x-intercept, y=0. But, we just decided that y can't be 0, so there is no x-intercept.

Hope that helps.

Aram

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