Question

What is the possible discriminant of the graph

Answer 1

https://vcs.brainhoney.com/Resource/4584903,6B,0,0,D/Assets/flvs/algebra2_v10_gs-xml/res0065/5_07g_03.gif

Answer 2

Do they mean what the x-intercepts are when y=0?

Answer 3

does not specify but they give me answers such as

-11
zero
18
36

Answer 4

Think about where the discriminant comes into the equation for zeroes.
\[ x=\frac{-b \pm \sqrt{\color{green}{b^2 - 4ac}}}{2a} \]If the discriminant is negative, then your zeroes are complex valued / not on the graph.
If the discriminant is 0, then you'd get +- 0, so the only solution is x=-b/2a.
If the discriminant is not a perfect square, then you'll have irrational solutions (you can't remove the root in the solution).
If the discriminant is a perfect square, then you'll have rational solutions (they're pretty nice numbers!)

Maybe that'll help you think about it.

Answer 5

* and your solutions / zeroes are (-4,0) and (2,0).

Answer 6

But none of the choices fit the answer ?

Answer 7

I guess they're talking about the y-value.

Answer 8

they're all possible values of the discriminant for an equation of the graph (y=ax^2 + bx + c)
D = b^2 - 4ac
the discriminant is part of the formula for finding your zeros of y=ax^2 + bx + c, so which of the values of the discriminant will get you values like -4 and 2?

IE:
\[ x = \frac{-b \pm \sqrt{D}}{2a};~~D=-11,~0,~18,~or~36\\\implies x=-4~and~x=2 \]

Answer 9

@AccessDenied thank you for explaining that. I though "discriminant" just meant the x-intercept(s).

Answer 10

@marco What all of this means is that you'll need to find the eqn of your parabola first.

http://www.youtube.com/watch?v=F_cnNZ0fCeQ

Then you'll be able to find your discriminant.

Answer 11

Or you could just define the equation itself and find the discriminant directly...
vertex = (-1,-9) / vertex form, or you could use zeroes x=-4 and x=2 w/ y=(x--4)(x-2) (just because we can look at the graph and see a=1 / the graph is a translation of x^2)

you don't necessarily have to define the equation itself, you can just use the properties to infer... although it might be more obvious w/ direct way. :P