Question 535714
{{{y=-x^2+6x+4}}}
Noticing that 6x is double the product of x and 3, I see the first two terms as part of a square. I add and subtract 9 (the square of 3), and group to get
{{{y=(-x^2+6x-9)+9+4}}}
I know that {{{-(x-3)^2=-(x^2-6x+9)=-x^2+6x-9}}} and just "completed the square."
That gets me to the equation of the parabola in vertex form:
{{{y=-(x-3)^2+13}}}
That form of the equation equation tells you the axis of symmetry of the parabola is {{{x=3}}}
and the vertex is at (3,13).
To get the vertex form, you could "complete the square" as I did, or you could try to memorize an unwieldy formula handed down by someone who just did the same work with the generic parabola equation {{{y=ax^2+bx+c}}}
to come up with the formula for the vertex form:
{{{y=a(x+b/2a)^2+(4ac-b^2)/4a}}}
In your case, your a, b, and c values were:
{{{a=-1}}}, {{{b=6}}}, and {{{c=4}}}
If you substitute a, b and c in the monster formula you get your vertex form.
Admittedly, there is an in-between way:
You could just remember the formula for the axis: {{{x=-b/2a}}},
which gives you the x-coordinate for the vertex, calculate the value to know what to subtract from x in the vertex form, and then calculate the y-coordinate for the vertex, which is also the last term of the vertex form, by substituting the x-coordinate for the vertex in the equation for the parabola