Finding range of rational function

Question

$$y=\frac {x^2+ ax-2}{x-a}$$

I have been told that the range of the function is set of all real values then I am told to find the set of values of a.

My attempt:

$(x-a)y=x^2+ax-2$

$x^2+x(a-y)+ ay-2=0$

Now putting D>=0 as I have assumed x belonging to real

$a^2 +y^2-6ay+8 \ge 0$

Now this equation is quadratic in $y$ and is positive therefore this should have discriminant less than $0$ as the graph will lie above x axis as leading coeffecient is positive .I get :

$36a^2-32-4a^2 \le 0$

Using this I got the set of values of a to be $[-1,1]$ but the values of a is given by $(-1,1)$......where am I doing wrong

One way of seeing the issue is that if numerator and denominator of original expression have a root same ($x=a$) here then the whole expression becomes linear for all $x\neq a$ and un defined for $x=a$. Numerator has root $x=a$ for $a=\pm 1$ — jonsno, Jul 02 '18 at 03:05
And that if $D=0$ in the quadratic in $x$ essentially means numerator has a root $x=a$ so this case is not to be excluded.. — jonsno, Jul 02 '18 at 03:18
@samjoe how does D=0 in the quadratic of x means root has a root of x=a — Hydrous Caperilla, Jul 02 '18 at 09:59

score 1 · Answer 1 · answered Jun 30 '18 at 11:06

1

Hint: the inequality is given by

$$(y-3a)^2+8(1-a^2)>0$$

answered Jun 30 '18 at 11:06

Dr. Sonnhard Graubner

95,283

I forgot to include but wouldn't the discriminant be greater than equal to zero – Hydrous Caperilla Jun 30 '18 at 11:49
I have written the discriminant for you, it depends on the parameter $a$! – Dr. Sonnhard Graubner Jun 30 '18 at 11:53
It depends on a alright but shouldn't the discriminant be greater than equal to zero – Hydrous Caperilla Jun 30 '18 at 14:12
for a = 1, y cant take the value 3 and for a = -1 , y cant take -3 . but it has to be all the real numbers as range. so they are excluded. cant take cause in the given form, denominator becomes 0. – mathemanics Mar 18 '20 at 21:48

Finding range of rational function

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