How do you find the range of #y = (x+2)/(x^2-9)# ?

1 Answer
George C.
Jan 2, 2018

Let #y = f(x)# and determine for which values of #y# there is a real solution #x#...

Explanation:

Given:

#y = (x+2)/(x^2-9)#

Multiply both sides by #x^2-9# to get:

#yx^2-9y = x+2#

Subtract #x+2# from both sides to get:

#yx^2-x-(9y+2) = 0#

This is in the form:

#ax^2+bx+c = 0#

with #a=y#, #b=-1#, #c=-(9y+2)#

Its discriminant #Delta# is given by the formula:

#Delta = b^2-4ac#

#color(white)(Delta) = (color(blue)(-1))^2-4(color(blue)(y))(color(blue)(-(9y+2)))#

#color(white)(Delta) = 1+4y(9y+2)#

#color(white)(Delta) = 36y^2+8y+1#

#color(white)(Delta) = 4/9(81y^2+18y+9/4)#

#color(white)(Delta) = 4/9((9y)^2+2(9y)+1+5/4)#

#color(white)(Delta) = 4/9((9y+1)^2+5/4)#

...which is always positive for any real value of #y#.

So the range is the whole of #RR#, i.e. #(-oo, oo)#

graph{(x+2)/(x^2-9) [-10, 10, -5, 5]}

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