Question

minimum or maximum of 2x^2+4x-18

Answer 1

it is a parabola that opens up (because the leading coefficient 2 is positive)
therefore it has a minimum, not a maximum

Answer 2

the minimum value is the second coordinate of the vertex
the first coordinate of the vertex of
\[y=ax^2+bx+c\] is \[-\frac{b}{2a}\]
you have \[y= 2x^2+4x-18\] so the first coordinate of the vertex is
\[-\frac{4}{2\times 2}=-1\]

Answer 3

the second coordinate of the vertex is what you get when you replace \(x\) by \(-1\)

Answer 4

\[y=a(x^2+bx/a ) + c\\
y=a(x+b/2a)^2+c-a(b/2a)^~completing ~square\\
y=f(x+b/2a) form, if~base ~is ~y=x^2,~then ~this~means~shift~b/2a~to ~left\\
x=-b/2a,x-coord~point~of~vertex\\
y=a(x+b/2a)^2 +c-a(b/2a)^2\]
y=f(x-k)+B form,
y=x^2+b, we know vertex at 0,b
similarily
for
y=f(x-k)+B, vertex at k,b
y=a(x+b/2a)^2 +c-a(b/2a)^2, vertex at (-b/2a,c-a(b/2a)^2)