how do I identify a parabola

Question

 

using y = x^2 - 4x -32 how do I identify if the parabola opens up or down so I can explain the conclusion.

 

 

Thank you

Answer 1

Hi L;

The x² is positive.  Therefore this opens up.

Answer 2

Why? Because of "end behavior". The behavior when |x| is large.

 

E.g., let |x| be a googol, i.e., x = ±10^100,

then y = (±10^100)^2 - 4(±10^100) -32

y = +10^200 ± 4(10^100) -32

y ≈ +10^200

So both the right and left sides shoot off to +∞ and the parabola opens up.

 

Notice that if the leading coefficient was negative the parabola's end behaviors would be down:

 

y = -2(±10^100)^2 - 4(±10^100) -32
y = -2*10^200 ± 4(10^100) -32
y ≈ -2*10^200

Answer 3

f(x) = aX^2 + bX + C     / if a>0 ,then parabola opens upward and vertex is at minimum point,

                                    / if a <0 , then parabola opens downward and vertex is at maximum point.