I want to find the inverse of: $y = -10 x^2 + 290 x - 1540$.
I tried using 'completing the square' to find it, but it did not work. I realize that the inverse will not be a function, but I still need this inverse.
If you want the complete question, here it is: The solar radiation varies throughout the day depending on the time you measure it. Say the radiation on a cloudy day can be approximated by the function $R(t) = - 10 t^2 + 290 t -1540$, where $R$ represents the solar radiation in $W/m^2$, and $t$ represents the time in hours from $0$ to $24$. What's the domain and range of the function? Find the inverse of this function. What does the inverse represent? Is the inverse a function?
Given a quadratic equation in the form
$$ y=ax^2+bx+c$$
it can always be re-written in the form
$$ y=a(x-h)^2+k\tag{1} $$
where $h$ is the value
$$h=-\dfrac{b}{2a} \tag{2}$$ which is the value of $x$ at the vertex of the graph.
We can find the value of $k$ by observing that it is the constant term when equation (1) is expanded. So
$$ ah^2+k=c \tag{3} $$
For your particular exercise, $a=-10, b=290, c=1540$. So from equation (2) we find that $h=14.5$ and from equation (3) we find that $(-10)(14.5)^2+k=-1540$. From this we conclude that $k=562.5$.
Substituting into equation (1) we get
$$ y=-10(x-14.5)^2+562.5 $$
To get the inverse relation we solve this equation for $x$ in terms of $y$.
\begin{eqnarray} -10(x-14.5)^2&=&y-562.5\\ (x-14.5)^2&=&\frac{562.5-y}{10}\\ x-14.5&=&\pm\sqrt{56.25-\frac{y}{10}}\\ x&=&14.5\pm\sqrt{56.25-\frac{y}{10}} \end{eqnarray}
Some instructors prefer that you exchange the $x$ an $y$ variables at this stage to get
$$ y=14.5\pm\sqrt{56.25-\frac{x}{10}} $$