Question

The parabola y=ax^2+bx has the range of y less than or equal to 0. Show that 16a+b^2 =0

Answer 1

if the range is \(y\leq 0\) this means the vertex is on the \(x\) axis

Answer 2

there is something wrong with this question
if the vertex is on the \(y\) axis, then you have a perfect square

Answer 3

Hmm, if \(a,b=0\) do you still have a parabola? I wouldn't think so.