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X and Y intercepts (amongst other things)?
I'm given 6 equations and asked for the domain, the x and y intercepts and to sketch a graph to indicate asymptotes along with finding equations for the asymptotes. Help with a couple examples? I'll post all 6 equations but I'm not really expecting anyone to show them all. If you could explain the process in one or two, that would be great.
a) f(x) = (2x)/(x-1)
b) g(x) = (3x+2)/(2x-5)
c) h(x) = (x+1)/(x-2)
d) j(x) = (4x-12)/(x+8)
e) k(x) = (8x+16)/(5x-0.5)
f) m(x) = (9x+24)/(35x-100)
1 Answer
- 1 decade ago
The domain would be all the possible values for x. In the first equation, x-1 is in the denominator so we don't want x - 1 = 0 so that means x can't be equal to 1. So the domain is {all real numbers excluding 1} By the way, you just found an asymptote also. The equation for the asymptote is x = 1
To find the intercepts, just substitute 0 in for x and that gives the y intercept and then substitute 0 in for y and that gives the x intercept. So, to find the y-intercept I solve f(0) = (0)/(0-1) = 0 This means that when x = 0, y = 0 so the graph crosses the x and y axis at (0,0)
Let's talk about the graph. When x approaches the asymptote ( x = 1) what will happen to the y value? (x - 1) is negative when x < 1 so f(x) will be negative also when 0<x<1. Since (x-1) is in the denominator, as x approaches 1, (x - 1) becomes close to 0 which means the fraction (2x)/(x-1) decreases to negative infinity the closer x gets to 1. So the graph curves down to infinity as x approaches 1. On the other side, x > 1 so the equation (x - 1) is positive thus f(x) is positive for x>1.
It also approaches positive infinity as x approaches 1 from the x>1 side. Ok, you pick some more points for x > 1 and graph them. Oh, one more thing, what happens when x < 0? Now the numerator and the denominator are negative so f(x) is positive again. Pick some points for x< 0 and graph those too. Thanks for the interesting problem, hope this helps. Have a blessed day.
