Is This Function a Variation of a Hyperbola?

In summary, the conversation discusses an equation and how to sketch its graph using a graphing calculator. The equation is y= (x-1)^2 / (x+1)(x-2). The conversation also mentions finding asymptotes and intercepts, and using methods such as polynomial division and limit calculus to accurately sketch the graph.
  • #1
dagg3r
67
0
hi i got this equation
y= x^2 - 2x + 1 / X^2 -x - 2

how do i sketch this finding all intercepts, and asymptotes with a gfx calculator? Please check if the steps i did below is right

what i did i factorised the equation so i got

y= (x-1)^2 / (x+1)(x-2)

ASYMPTOTES
the bottom line (x+1)(x-2) = 0 gives me the vertical asymptotes thus
x= -1 and x=2

X-Intercept
to find the x-intercept i let y=0 thus (x-1)^2
X=1

Y-Intercept Let X=0
y=-1/2

I got those values but how do i sketch the graph it looks weird on the gfx calc is this a variation of a hyperbola? i also heard you can do stuff like use polydivion and get another equation or break it up and sketch it or something like that?
 
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  • #2
dagg3r said:
I got those values but how do i sketch the graph it looks weird on the gfx calc is this a variation of a hyperbola? i also heard you can do stuff like use polydivion and get another equation or break it up and sketch it or something like that?

How precise does the graph need to be? If I were doing this for a class, I would just calculate all of the interesting points and trends and then connect the dots. For example, what happens as x goes to inf and -inf? Does y go to +inf or -inf at the critical points? I'm not sure why it won't plot on your calculator, though. Do the infinities cause problems? If so, have you tried restricting the range?
 
  • #3
The horizontal asymtote needs to be calculated as well.

To do this, I am going to use limit calculus.

[tex]
y = \lim_{x\rightarrow\infty} \frac{x^2 - 2x + 1}{x^2 - x - 2}
[/tex]
[tex]
y = \lim_{x\rightarrow\infty} \frac{\frac{x^2}{x^2} - \frac{2x}{x^2} + \frac{1}{x^2}}{\frac{x^2}{x^2} - \frac{x}{x^2} - \frac{2}{x^2}}
[/tex]
[tex]
y = \lim_{x\rightarrow\infty} \frac{1 - \frac{2}{x} + \frac{1}{x^2}}{1 - \frac{1}{x} - \frac{2}{x^2}}
[/tex]
[tex]
y = \frac{1 - 0 + 0}{1 - 0 - 0}
[/tex]
y = 1

Another way to do this would be to switch x and y, solve for y, and see which values of x are undefined for that function, but the calculus method is quicker and easier IMO.
 
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FAQ: Is This Function a Variation of a Hyperbola?

1. What is the equation of a hyperbola?

The general equation of a hyperbola is x^2 / a^2 - y^2 / b^2 = 1, where a and b are the distances from the center to the vertices in the x and y directions, respectively.

2. How is a hyperbola different from a parabola?

While both hyperbolas and parabolas are conic sections, the main difference between them is that a hyperbola has two separate branches, while a parabola has only one. Additionally, a hyperbola has two asymptotes, while a parabola has none.

3. What is the significance of the eccentricity in a hyperbola?

The eccentricity of a hyperbola is a measure of how elongated the hyperbola is. It is defined as the ratio of the distance between the foci to the length of the major axis. A hyperbola with an eccentricity greater than 1 is called a "stretched" hyperbola, while a hyperbola with an eccentricity less than 1 is called a "squeezed" hyperbola.

4. How does the orientation of a hyperbola affect its equation?

The orientation of a hyperbola refers to the direction of its opening. A hyperbola can open horizontally (with its branches extending left and right) or vertically (with its branches extending up and down). The orientation affects the placement of the x and y terms in the general equation, but the rest of the equation remains the same.

5. Can a hyperbola have a center?

No, a hyperbola does not have a center. Unlike a circle or ellipse, which have a defined center point, a hyperbola has two branches that approach but never intersect. This means that there is no single point that can be considered the center of a hyperbola.

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