Stuck on Three Problems Please Help :)

Please post one question at a time.

3. What is the focus of the parabola?

y=−14x2−x+3

Enter your answer in the boxes.

$y=−14x^2−x+3\\ y-3=−14(x^2+\frac{1}{14}x)\\ \frac{-1}{14}(y-3)=x^2+\frac{1}{14}x\\ \frac{-1}{14}(y-3)+\frac{1}{784}=x^2+\frac{1}{14}x+\frac{1}{784}\\ \frac{-1}{14}(y-3-\frac{1}{56})=(x+\frac{1}{28})^2\\ (x+\frac{1}{28})^2=\frac{-1}{14}(y-3\frac{1}{56})\\ (x+\frac{1}{28})^2=4*\frac{-1}{56}(y-3\frac{1}{56})\\$

concave down parabola

$Vertex=(\frac{-1}{28},3\frac{1}{56})\\ Axis\; of \;symmetry:\quad x=\frac{-1}{28}\\ focal\;length=\frac{1}{56}\\ focus=(\frac{-1}{28},3\frac{1}{56}-\frac{1}{56})\\ focus=(\frac{-1}{28},3)\\ $

Here is the graph

The others must be done by rearranging the equations as well.

you need them of the form

$(x-h)^2=4a(y-k)$

Where (h,k) is the vertex and |a| is the focal length.

The last one is sideways so the x and y will be swapped around.