Given: #y = x^2 - x + 5#

The domain of an equation is usually #(-oo, oo)# or all reals unless there is a radical (square root) or a denominator (causes asymptotes or holes).

Since this equation is a quadratic (parabola), you would need to find the vertex. The vertex's #y#-value will be the minimum range or the maximum range if the equation is an inverted parabola (when the leading coefficient is negative).

If the equation is in the form: #Ax^2 +Bx +C = 0# you can find the vertex:

vertex: #(-B/(2A), f(-B/(2A)))#

For the given equation: #A = 1, B = -1, C = 5#

#-B/(2A) = 1/2#

# f(1/2) = (1/2)^2 - 1/2 + 5#

# f(1/2) = 1/4 - 2/4 + 20/4#

#f(1/2) = 19/4 = 4.75#

Domain: #(-oo, oo)# or all reals

Range: #[19/4, oo)# or #" "y >= 19/4#

graph{x^2-x+5 [-25.66, 25.66, -12.82, 12.83]}