hi can someone help me with this problem where i need to find the range for the equation: y=(2/x+6)
let's us notice first that the domain, that is, the values "x" can take, must not make the denominator =0, so what values "x" MUST NOT take?
i mean -6
x cannot =-6 but how does that help with range, that only helps with finding the domain i think
so... yes... x cannont be -6, or it makes the denominator zero other than that, "x" can take any positive vlaue or any negative value so let's see what happens when say we give it a few values, let's see what happens to "y"
ok
gimme a sec
ok
hmm, one sec
ok
\(\large y = \cfrac{2}{x+6}\\ \quad \\ \quad \\ \begin{array}{llllllll} x = 6&y = \cfrac{2}{x+6}& y = \cfrac{1}{6} \\ x = 5&y = \cfrac{2}{x+6}& y = \cfrac{2}{11}\\ x = 4&y = \cfrac{2}{x+6}& y = \cfrac{1}{5}\\ x = 3&y = \cfrac{2}{x+6}& y = \cfrac{2}{9}\\ x = 2&y = \cfrac{2}{x+6}& y = \cfrac{1}{4}\\ x = 1&y = \cfrac{2}{x+6}& y = \cfrac{2}{7}\\ x = 0&y = \cfrac{2}{x+6}& y = \cfrac{1}{3}\\ x = -1&y = \cfrac{2}{x+6}& y = \cfrac{2}{5}\\ x= -2&y = \cfrac{2}{x+6}& y = \cfrac{1}{2}\\ x = -3&y = \cfrac{2}{x+6}& y = \cfrac{2}{3}\\ x = -4&y = \cfrac{2}{x+6}& y = 1\\ x = -5&y = \cfrac{2}{x+6}& y = 2\\ x = -7&y = \cfrac{2}{x+6}& y = -2\\ x = -8&y = \cfrac{2}{x+6}& y = -1\\ x = -9&y = \cfrac{2}{x+6}& y = -\cfrac{2}{3}\\ \end{array}\)
okay so what does that mean
?
so notice how far up the range goes, at x = -5 "y" is 2 x = -7 "y" goes down to -2 if it goes to the right of -5, is a fraction if it goes to the left of -7, is a fraction too a fraction is less than 1
range is the value of "y"
so.... what do you think would be the interval for the range, that is, for the value of "y"?
i have no idea?!
depending on values taken by "x", valid values, not like -6 "y" will take its value the range is how far up and down "y" goes
i knew what range means but i dont get what would be the interval for range?
at x = -5 is how far UP it goes, +2 and at x = -7, is how far DOWN it goes -2 so the range would be [-2, 2]
okay i get it but isnt there a simpler way to do it i dont have to substitute all these values of x to find y right?
well, the other way would be graphing it
right i did that but i couldnt see the range on my calculator how do i find it on the calculator?
hmm... actually, the graphs shows ... a different range http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIyLyh4KzYpIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTEyLjYyIiwiMC4zNzk5OTk5OTk5OTk5OTk0NSIsIi0xLjA0MDAwMDAwMDAwMDAwMDUiLCI2Ljk1OTk5OTk5OTk5OTk5OSJdfV0-
on that graph what would the range be?
the range is the values "y" takes, the vertical lines
right so would it be y is greater than or equal to 0, and y is less than or equal to 0?
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