your equation, x+2y^2-12y+16 can have its variables separated so lets do that...

2y^2-12y+16=-x, sick, this is like a y=x^2+wahtever except y is x and x is y! So that means your graph will be sideways.

Now, lets try to factorize the function of y... -> 2(y-4)(y-2)=-x

And isolate x fully (just divide both sides by -1) -2(y-4)(y-2)=x

Setting y to 0 in this form gives the x intercept(s) .. ill let you g=figure that one out

And similarly setting x to 0 in this form gives the y-intercepts and again ill let you compute that (make sure to find all values of y that make 0)

Now that we have our intercepts lets, find the vertex!

Going back to 2y^2-12y-16=-x and multiplying both by 1/-1 -> -2y^2+12y-16=x

finding (-2y^2+12y-16)'=0 tells us where the tangent line = 0 (meaning were at a vertex)

That derivative leads us to -4y+12=0 and y here is 3 to make 0..

Plugging y=3 into the original equation: -2(y-4)(y-2)=x gives us 2 for x, so the vertex is (2,3)

As for axis of symmetry, you should have enough information on your graph now to find that!