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(1 point) Consider the function $f(x) = \frac{-2x + 5}{8x + 4}$ Enter the equations of the vertical asymptotes. If there are no vertical asymptotes, enter none. If there is more than one vertical asymptote, enter a list of the equations separated by commas (e.g., $x = 20$, $x = -7$). Vertical asymptotes: Find the $x$-intercept(s). If there is more than one $x$-intercept give a list of the $x$-intercepts separated by commas (i.e.: $(1, 2), (3, 4)$). If there is no $x$-intercept type in none. $x$-intercepts: Find the $y$-intercept: Find the domain of $f(x)$: Give your answer in interval notation.

Transcript

0:00 Hi there.

00:01 In this question, we are given this function, f of x is equal to 1 over 1 minus x squared.

00:07 So in part a, we have to find the domain of the function.

00:12 So the domain would be all real numbers, but this is a fractional number.

00:16 So the domain, i mean, the denominator cannot be zero.

00:19 So 1 minus x square cannot be 0.

00:22 That means x square cannot be 1.

00:24 That means x cannot be plus or minus negative 1.

00:28 So the domain of the function, which is all real numbers except negative 1 and 1.

00:36 So this is the domain.

00:38 And x intercept, for the x intercept, we have to just set this equation equal to 0, which is 1 over 1 minus x square.

00:48 So if i just do the cross multiplication, 0 is equal to 1.

00:51 So this is not equal.

00:53 That means no x intercept.

00:58 And for y the y intercept, so i have to find the f of 6.

01:03 0 which is 1 over 1 minus 0 squared which is 1 so the y intercept is the point where it is it has the coordinates 0 n 1 and so is it an even function or not so for even and for the odd functions we have to just find f of negative x 1 minus 1 minus x squared which is equal to 1 over 1 minus x squared which is equal to f of x.

01:31 So since f of negative x is equal to f of x, this is an even function.

01:40 So all these answers are for part a.

01:45 And for part b, find the vertical and horizontal asymptotes.

01:49 For the vertical asymptotes, we are going to just make the denominator equal to zero.

01:54 That means x is equal to plus or minus one.

01:57 So we have two vertical asymptotes.

01:59 The first one is x is equal to 1.

02:01 The next one is x is equal to negative 1.

02:04 And for the horizontal asymptote, we have to just take the limit of the function...