Graphing Question....Algebra 2.?

Question

I have a question about this equation:

y= 5/x^2+1

So, there isn't an x intercept and the horizontal asymptote is y=0 since p(x) degree < q(x) degree.
And isn't the denominator supposed to be the vertical asymptote? So, the vertical asymptote would be - 1, wouldn't it? But when I checked the graph on my graphing calculator, there wasn't a vertical asymptote. Why is that?

Answer 1

Vertical asymptotes of a rational function can only occur when the denominator is 0. But x²+1 is never 0 for any
real x, so there is no vertical asymptote.

Answer 2

The slope-intercept type is: y = mx + b the position m is the slope of the line (substitute in y values for each substitute in x values), and b is the y-intercept (the position the line crosses the y axis, at the same time as x=0). factors on the line are frequently given as (x, y). commonly the finest thanks to graph on paper a given line, is to commence off with the intercept factors, (0, y-intercept) and (x-intercept, 0). Assuming that you do have a line contained in the above slope-intercept type, then you actually can purely get a ruler and draw a line by the above 2 factors to furnish you your graphed line. or purely plug in a fee for both x or y and sparkling up to discover the point, and upon getting 2 or extra of those factors you could graph/draw the line. So, for y = 4x - 2, m=4 and b = -2. you've already got your first element: your y-intercept is (0, -2). in case you pick to apply the slope m of four, then you actually can use the 4:a million ratio of y:x to characteristic for your y-intercept element to discover the subsequent element: (0+a million, -2+4), which provides element (a million,2). in case you pick the x-intercept, set y=0 and sparkling up for x: 0 = 4x - 2 0+2 = 4x -2 + 2 2 = 4x 2/4 = 4x/4 a million/2 = x so the x-intercept is (a million/2, 0) You responded your self for what the y-intercept is for y = (5/3)x. Rewrite into the slope-intercept type to make sure that: y = (5/3)x + 0 so there's a y-intercept, although that is at (0,0) and also you could purely proceed graphing from there.

Answer 3

x^2+1 is never zero because x^2 doesn't equal -1 for real values of x.