I was given an equation to analyze and graph as a final requirement to pass but then again I have no idea how to.

[James Ibarra]

1. Analyze completely and graph the curve represented by the equation y²=(x²-2x)/(x-1)

Since i was hospitalized for almost 3 weeks, I missed most of the online lessons my class instructor gave from the past month. But i was still needed to pass something in order to pass the subject. I was given this problem but i missed most of the lessons and I really don't know how to tackle this. is this a bit close graphing linear equations like y = mx +b? or is it far from it?

 

[Dr.Peterson]

I can't tell what "analyze completely" means without knowing your context. What course are you taking, and what were the lessons about, that this is testing?

If it weren't for the square on the left, this would be about the techniques for graphing a rational equation, which are well-known. With the square, it strikes me as a little beyond a basic calculus class (as it isn't even a function, though the techniques there could be used).

In any case, it is far beyond graphing linear equations.

What topics did you miss, and what did you not miss?

 

[James Ibarra]

I can't tell what "analyze completely" means without knowing your context. What course are you taking, and what were the lessons about, that this is testing?

If it weren't for the square on the left, this would be about the techniques for graphing a rational equation, which are well-known. With the square, it strikes me as a little beyond a basic calculus class (as it isn't even a function, though the techniques there could be used).

In any case, it is far beyond graphing linear equations.

What topics did you miss, and what did you not miss?

I missed the conic sections topic completely and only got up into triangles and trigonometric identities, i was given this problem as a final requirement. I graphed it into desmos quadratic equation/parabola grapher and got this graph haha

 

[Dr.Peterson]

I missed the conic sections topic completely and only got up into triangles and trigonometric identities, i was given this problem as a final requirement. I graphed it into desmos quadratic equation/parabola grapher and got this graph haha

This is not a conic section, and involves no trigonometry. If your course is a typical precalculus course, which it sounds like though you haven't given me enough information, I can't think where this equation would fit in. So I still can't tell what "analyze completely" means in your context, even if you were just supposed to graph it with technology and describe the features you see.

If I were trying to help you face to face, I would have grabbed your textbook and syllabus to see what it covers that would relate to this problem. If you skim the sections you missed, do you see anything at all like this?

 

[James Ibarra]

This is not a conic section, and involves no trigonometry. If your course is a typical precalculus course, which it sounds like though you haven't given me enough information, I can't think where this equation would fit in. So I still can't tell what "analyze completely" means in your context, even if you were just supposed to graph it with technology and describe the features you see.

If I were trying to help you face to face, I would have grabbed your textbook and syllabus to see what it covers that would relate to this problem. If you skim the sections you missed, do you see anything at all like this?

I am taking pre calculus grade 11 course, I think I will just get every value i can. (x and y intercepts, asymptotes). also we do not use books anymore, I asked my instructor he said that i should understand this equation. If i cannot get any value of move past this current equation I think I'll just ask my instructor to swap this problem to a set of problems regarding more on shapes(conic sections)

Our last topic was just conic sections. the shapes circle, ellipse, parabola, and the hyperbola and nothing more. here is the pdf

 

[Dr.Peterson]

I think I will just get every value i can. (x and y intercepts, asymptotes). also we do not use books anymore, I asked my instructor he said that i should understand this equation.

I agree, just find everything you can, using whatever tools you know. Missing the section on conic sections should not affect what you can do.

1. Analyze completely and graph the curve represented by the equation y²=(x²-2x)/(x-1)

You might first analyze y=(x²-2x)/(x-1), which is a rational function that you can probably handle, and then consider the effects of taking the square roots of that (which accounts for the symmetry across the x-axis, and affects the shape of the x-intercepts as well as the oblique asymptote of the rational function).

 

[topsquark]

Some random thoughts. Some of these may not be possible to calculate but it'll give you some idea of what you might be expected to look for.

Given y = f(x).
1) Find the roots, where f(x) crosses the x-axis, ie. solve 0 = f(x).

2) Find where the graph crosses the y-axis, ie. solve y = f(0). (There will be only one of these for a function.)

3) Look for asymptotes:
a) Horizontal asymptotes. Find [math]\lim_{x \to \pm \infty} f(x)[/math].
b) Vertical asymptotes. These generally involve functions of the form [math]f(x) = \frac{g(x)}{h(x)}[/math]. Look for where h(x) = 0.

4) Look for slant asymptotes. These are a bit harder to find. If you are dealing with rational functions of polynomials [math]f(x) = \frac{g(x)}{h(x)}[/math] where the degree of g(x) is one more than the degree of h(x) you will have a slant asymptote of the form mx + b. You will need to know how to polynomial division to find these.

5) Look for places that f(x) does not exist. For example, [math]y = \sqrt{x}[/math] does not exist for any [math]x < 0[/math] and [math]y = ln(x)[/math] does not exist for [math]x \leq 0[/math].

There may be more that your instructor will be looking for but it'll get you started.

Note: If you are looking at a Calculus treatment there is much more you can find out. I've listed some ideas that only come from a non-Calculus format.

-Dan

 

[HallsofIvy]

Looking at the right side only, \(\displaystyle \frac{x^2- 2x}{x- 1}= \frac{x(x-2)}{x-1}\) has zeros at x= 0 and x= 2 and a vertical asymptote at x= 1. But that is \(\displaystyle y^2\). \(\displaystyle y= \pm\sqrt{\frac{x(x-2)}{x-1}\). The square root of 0 is still 0 and the square root of "infinity" is still "infinity" so that will still be 0 at x= 0 and x= 2 and still have a vertical asymptote at x= 1 but it will be "mirrored" in the x-axis.