A function does not have any x-intercepts. What
might be true about its domain and range?

The domain exists on all real numbers i.e {x∈R}∈

The range exists all on real numbers except at y = 0

Domains are all input values of a function for which the function exists while ranges are all the output values for which the function exists.

Since the x-intercept of a function exists at where y = 0, this means that the point where a function does not have any x-intercepts are all other points on the graph except at y = 0.

The following statements are therefore true;

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Here's my two cents worth about this one:

This is a discontinuous function - and it passes the vetical line test because any vertical line would only cut the "graph" in one place. The graph would have an infinite number of disconnected "peaks" and "valleys" because the x values along the number line are continuous. In other words, all the rationals would have y values of 1 and all the irrationals would have values of 0. The graph is definitely non-continuous because there aren't any "intermediate"  y values - i.e., "y" is either 0 or 1 - and nothing in between !!   Basically, the graph would be just  "dots" at either y= 0 or at y =1. In a strange way though, this is an "even" function, because the negative of any positive rational would have a corresponding y value of 1 and the negative of any positive irrational would have a corresponding y value of 0. Thus, f(x) = f(-x). The y intercept would be  the "dot" at (0, 1) - because 0 is rational. The "x" intercepts would just be the "dots" where the irrational numbers are located on the number line because "f(x)" would be 0 at those points!!!  The domain is all real numbers, but there are onl two values for the "range" 0 and 1 !!!

That's my "take" on this one !!!

Step-by-step explanation:

michael scott & dwight schrute

step-by-step explanation:

lazy scranton music video

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5