math problem hard to solve

Question

Determine if the following functions are one-to-one, even,odd, periodic.

(a) f(x)=x²+6x−8, for x∈[3,∞)

please tell me I don't get this tell the right answer so I can understand

Answer 1

x^2 + 6x - 8 = 0

x^2 +6x + 3^2 = 8 + 3^2

(x-3)^2 = 17

y= (x-3)^2 -17 is an upward opening parabola with vertex and minimum point at (3, -17). It's line of symmetry is x=3

for x> or = 3, the function is one to one. It passes the vertical line test and the horizontal line test. It's impossible to draw a vertical or horizontal line that cuts the graph more than once. That makes it one to one

That graph is not even or odd or periodic.

It doesn't repeat, so it's not periodic

f(x) does not equal f(-x) or -f(x) so it's not even or odd

x^2 + 6x - 8 does not equal (-x)^2 +6(-x) - 8

or -[(-x)^2 +6(-x) -8]