Equations - Quadratic equations and graphs #1

Question

callisto · Answer

Perhaps this is not difficult, I'm not sure, but I can't get the answer :S

ash2326 · Answer

@Callisto 
we have
$$y=ax^2+bx+c$$
In the figure, we notice that parabola is opening up
$$=> a\ is\ postive$$

callisto · Answer

Do you get B as well?

ash2326 · Answer

Another important thing to notice is that the parabola can never intersect x-axis or it has no real zeros

anonymous · Answer

B, it is.

ash2326 · Answer

This implies
$$=> Discriminant\ is\ negative$$

ash2326 · Answer

or 
$$b^2-4ac<0$$

anonymous · Answer

B.

callisto · Answer

I got B too. But the answer is D. Then I was thinking... are those all I've learnt wrong?!

anonymous · Answer

If a was less than 0, the graph would have a max not a min.

callisto · Answer

c is the y-int , so c>0

ash2326 · Answer

Now we have to find the sign of c $$ b^2-4ac<0$$ $$ b^2-\underline{4ac}<0$$ a>0 so c>0 to make 4ac>0 so it's b

callisto · Answer

So, I just want to make sure that I got the correct one. Thanks! especially a big thanks to @ash2326 (Actually, I knew them all... sorry for causing you so much time to explain)

ash2326 · Answer

No worries, I helped you=>I helped myself

anonymous · Answer

So Callisto's been using us :o

callisto · Answer

Using? Why?

anonymous · Answer

I was trolling hehe ...Sorry. :D

callisto · Answer

Oh... no sorry :)