If a vertical parabola opens upward, has its vertex in the third quadrant, and $y=ax^2+bx+c$ is the equation of this parabola, which of the following can be true? Sketch a curve for each possible case.
$a>0$, $b>0$, $c>2$
I think it could be both ways. I know $a$ and $b$ are $> 0$ but $c$ (or $y$-intercept ) could lie below the origin or above it. Please help.
The fact that the parabola opens upwards forces $a>0$. The $x$-coordinate of the vertex of the parabola is $-\frac{b}{2a}$; since $a>0$ and this expression is negative, we have $b>0$ as well.
Finally, the $y$-coordinate of the vertex of the parabola is $$a\left(-\frac{b}{2a}\right)^2 = b\left(-\frac{b}{2a}\right) + c = c-\frac{b^2}{4a}.$$ This expression must be less than zero. But this doesn't place much of a restriction on $c$. For example, $a=b=1$ implies that $c<\frac{1}{4}$, but it can be positive or negative. $a=1$ and $b=4$ implies that $c < 4$, so we could have $c>2$.