$x$ and $y$ are real numbers, $x^2-2x-4y=5$, what is the range of $x-2y$? $x$ and $y$ are real numbers, $x^2-2x-4y=5$, what is the range of $x-2y$?
I think I'm supposed to rewrite the equation in terms of $(x-2y)$ and get a quadratic, but I'm not sure if I'm really supposed to do that. I tried completing the square with $(x-1)^2$ and got: $(x-1)^2=6+4y$ but didn't know what to do after that. I can get the minimum value of $6+4y$ but that really doesn't help me at all. Please help me out!
 A: Let's call $u=x-2y$. Then you can rewrite your equation as $$x^2-2x-4\frac{x-u}2=5$$
or $$x^2-4x+(2u-5)=0$$
You want $u$ such that the equation above has a real solution, so $$2^2-(2u-5)\ge 0$$ or $$u\le\frac 92$$
A: Your idea of using $(x-1)^2=6+4y$ was good.
In terms of a parameter $t$ we can write $$x=1+t, y=\frac{t^2-6}{4}.$$
Then  $x-2y=\frac {9-(t-1)^2}{2}$.
The range is $(-\infty,\frac{9}{2}]$.
A: $x^2-2x-4y=5$ is equivalent to $y=\frac{1}{4}(x^2-2x-5)$. Putting this into $x-2y$ gives us $\frac{9}{2}-\frac{1}{2}(x-2)^2$. The range will be $]-\infty,\frac{9}{2}]$. You'll need to explain why though. Intuitively, one can choose $x$ freely, hence the term $\frac{1}{2}(x-2)^2$ can be any non-negative real number.
A: From our equation we get
$$y=\frac{1}{4}(x^2-2x-5)$$ so we get
$$x-2y=x-\frac{1}{2}(x^2-2x-5)=-\frac{1}{2}x^2+2x+\frac{5}{2}$$
A: Let's see: $(x-2y)^2 = x^2 -4xy + 4y^2,$ introducing terms in $xy$ and $y^2$ that are not in the original equation. I think I would not take that any further.
One thing you can try is to find $y$ in terms of $x$ and substitute into $x-2y,$
so now you just need to find the range in terms of $x.$
Of course you must also check for restrictions on $x$ in the given equations; are there any?
A: $$x^2-2x-4y=5 \implies y=(x^2-2x-5)/4$$
$$x-2y = \frac {-1}{2}[(x-2)^2-9]$$
Thus the range of $x-2y$ is $(-\infty, 4.5]$
A: From your result $(x-1)^2=6+4y,$ we get $y=\dfrac{(x-1)^2}4-\dfrac32,$
so $x-2y=x-\dfrac{(x-1)^2}2+3=-\dfrac{x^2}2+2x+\dfrac52=-\dfrac12(x^2-4x-5)$
$=-\dfrac12\left(\left(x-2\right)^2-9\right)\le\dfrac92.$
A: 1) $y=(1/4)(x-1)^2- 3/2$, a parabola opening upwards, vertex $(1,-3/2)$.
2) $x-2y=C$, a line ;
$y=(1/2)x -C/2$;
Looking for the range of $y-$intercept $-C/2$ of the family lines that intersect the parabola.
Slope of tangent to parabola at point of tangency with line equals $1/2$:
$y'=(1/2)(x-1)=1/2$; Then $x=2$;
Line $y=(1/2)x -C/2$ passes through $(2,-5/4)$.
$-5/4=(1/2)2-C/2$;
$C=9/2$;
For $-C/2 >-9/4$, i e. $C<9/2$ the line intersects the parabola.
Range of $x-2y$:  $(-\infty, 9/2]$
