Finding features of parabolic formula I have a question that requires me to find the intercepts and the turning point of a parabola from it's formula, but I'm not really sure how to find all this information out from the equation below. I tried using completing the square but I ended up with a weird answer that I don't believe is right. The equation is:
$$y=-3x^2-5x+2$$
Thanks!!

There are two intercepts. The $y$-intercept, and the $x$-intercepts.
$y$-intercept:
To find the $y$-intercept, you simply plug in $x=0$ into the equation to get the $y$ value. So thus, we have$$y=-3\cdot 0^2-5\cdot 0+2=2$$ so the $y$-intercept is $(0,2)$.
$x$-intercepts:
The $x$ intercepts are points that give $y=0$ when you plug in $y$. So plugging in $y=0$, we get $$-3x^2-5x+2=0\Leftrightarrow -(3x-1)(x+2)=0\implies x=\frac 13,x=-2$$
Therefore, the $x$-intercepts are $\left(\frac 13,0\right)$ and $(-2,0)$.

To find the Vertex, you can substitute $x$ with $\frac {-b}{2a}$ where given the quadratic of the form $ax^2+bx+c=0$.
In your example, the quadratic is $y=-3x^2-5x+2$ so we see that $a=-3,b=-5,c=2$. So we have $$-\frac b{2a}=-\frac 56\tag1$$
Substituting $x$ with $(1)$ gives the $y$ value:$$f\left(\frac 56\right)=-3\left(-\frac 56\right)^2+5\left(\frac 56\right)+2=\frac {49}{12}$$
And so the vertex lies on the point $\left(-\frac 56,\frac {49}{12}\right)$.
