Question 42723
Well from
g(x) = x^2 - 2x - 15,
we can find the vertex by looking at -b/2a for the x-coordinate and g(-b/2a) for the y-coordinate...thus we get
-b/2a = 1 and
g(1) = -16
so our vertex is at (1, -16)
Thus our parabola is of standard width, concave up, vertex at (1, -16) and has x-intercepts at (5, 0) and (-3, 0), which can be found by factoring g(x)...
Its domain is all real numbers and the range would be g(x) ≥ -16