These are the steps :
1. find x and y intercept
2. find vertical and horizontal asymptotes
3. find stationary points
4. find the intersection between horizontal asymptote and the graph
1. I'm sure you can find the x and y intercept ( intersection between the x and y axis)
2. To find vertical asymptote : set the denominator = 0 and solve for x
3. To find horizontal asymptote : horizontal asymptote = y =
4. To find stationary points : set and solve for x. Then subs. the value of x obtained to f(x) to get the coordinate of stationary point(s).
5. To find the intersection between horizontal asymptote and the graph, just substitute the value of horizontal asymptote to f(x) and solve for x to get the coorodinate
I haven't done the question yet because I've been watching some math videos as I'm still quite lost, but in this chapter of my textbook it doesn't mentioned what you've said - the textbook really isn't good though. Still not sure what to do... the most it mentions is the horizontal and vertical asymptote and that's it. The instructions are terrible and the examples in the textbook doesn't really help me with this question.
Hi eXist and marie7
Sorry I'm not aware this is pre-calculus sub forum.
@eXist : maybe there is another way to find the stationary points?
1. Intersection with the x-axis means that the value of y = 0, so
Hence, the x-intercept = (3,0)
2. Intersection with y-axis means that the value of x = 0, so
Hence, the y-intercept = (0,3)
Have you been taught about asymptotes, marie? ^^
Hi there Songoku,
Thanks for your help by the way!
No, I'm not sure how to handle asymptotes. I also need to graph the function. Could you show me the steps? That's all the question is asking, for me to graph the function. Thanks so much
The zeroes are at and
So you find the mid point between (2, 0) and (-2, 0). The x value that is, which comes out to be .
Then you substitute this value back into the original equation to get (0, 4). So therefor, the stationary point for is (0, 4).
y = 0 when x = 3 therefore, y intercept at x = 3
x does not = 0 at any point, therefore horizontal asymptote at y = 0
x cannot equal positive or negative 1, as this would mean the slop of the curve is undefined at these points, therefore vertical asymptote at x = -1 and +1
the curve is convex from -1<x<1, and concave for x<-1 and x>1
Is this right? And if so, can someone please show me what the graph is supposed to look like? just to check. because i'm still a little confused. This is a rectangular hyperbola, am I right? Thank you.