# Domain and range of $\sqrt {x^2-2x-8}$

Find the domain and range of: $$y=\sqrt {x^2-2x-8}$$

My Attempt: For domain, $y$ is defined iff $$x^2-2x-8\geq 0$$ $$(x-4)(x+2)\geq 0$$ $$x\in (-\infty, -2]\cup [4, \infty)$$

For Range, $$y=\sqrt {x^2-2x-8}$$ $$y=\sqrt {(x-1)^2 - 9}$$ $$y^2=(x-1)^2-9$$ $$y^2+9=(x-1)^2$$

• Shouldn't the range be y =>0 as this is in the real plane? – Mohammad Zuhair Khan Dec 26 '17 at 1:54
• Technically I am unsure what are we supposed to do. Anyway what you say is true enough. – Mohammad Zuhair Khan Dec 26 '17 at 1:57
• What is the question here? – Dave Dec 26 '17 at 1:58
• @Dave, what's the range of the function? – pi-π Dec 26 '17 at 2:00
• You've found the domain. One way to find the range is to find the minimum of the function on its domain (call this point $a$). Then, since the function is clearly unbounded on this domain, the range will be $[f(a),\infty)$, since the range is certainly a subset of $[0,\infty)$. – Dave Dec 26 '17 at 2:03

When you squared $y$ you introduced spurious solutions. Your work on the domain shows $y$ can go to $0$ and when $x$ gets large in either direction $y$ gets large and positive. What does that tell you about the range?
• Your final equation has a solution $(6,-4)$ which the original equation does not. That might convince you that $-4$ is in the range, but the original requires $y \ge 0$ – Ross Millikan Dec 26 '17 at 2:07
Notice that the term inside of the square root goes from $0$ to $+\infty$ in either direction. Therefore, $y$ would also go from $0$ to $\infty$