domain of 16x^2+8x-1?
if you meant
y = 16x^2 + 8x - 1
x of the vertex is -8/32 = -1/4
y = 16(1/16) + 8(-1/4) - 1
= -2
vertex is (-1/4), -2) and the parabola opens up
the domain of any "vertical" parabola is
x is any real number
confirmation:
http://www.wolframalpha.com/input/?i=y+%3D+16x%5E2+%2B+8x+-+1
domain of 16x+5
To find the domain of a function, you need to determine the values of x for which the function is defined. In this case, you have a quadratic function: f(x) = 16x^2 + 8x - 1.
Quadratic functions are defined for all real numbers, so the domain is all real numbers, unless there are any restrictions present in the problem. In this case, there are no such restrictions mentioned.
Hence, the domain of the function f(x) = 16x^2 + 8x - 1 is all real numbers (-∞, +∞).