Question

find the vertex of y=x^2-8x+19

Answer 1

The x coordinate of the vertex is -b/2a
Calculate that and plug it back in to find the y coordinate of the vertex.

Answer 2

\[y=ax^2+bx+c\]

Answer 3

There is a better way of thinking it that does not involve using formulas, but an understanding of the function you are dealing with.
In this case the function is a parabola:Now we can see that always for a given x there is another x that gives you the same y, and those x are simetrical to the vertex, therefore if you calculate both x for a given y the everage of the x found will be the x of the vertex.
In this case we can try to make it simpler by making the equation into a complete square, so we choose y=3, and we solve\[0=x^2-8x+19-3=(x-4)^2\]This means that the value that gives y= 3 for the the equations are both x=4 and by pure chance we see that the value is the one we are looking for since they are the same and that can only happen in the vertex.
In the general case this is the line of thought used to get to the formula showed above by @Mertsj and to show that you do the following:\[y(x)=ax^2+bx+c\]We then solve the equation of the given y:\[0=ax^2+bx+c-y\]We now multiply everything by 4a in order to make a perfect square again:\[0=4a^2x^2+4abx+4a(c-y)\]The first term we now see that is (2ax)^2 and the second is 2(2ax)(b) now the only thing that is missing so that we have a perfect square is b^2 so we sum it on both sides:\[b^2=4a^2x^2+4abx+b^2+4a(c-y)=(2ax+b)^2+4a(c-y)\]\[b^2-4a(c-y)=(2ax+b)^2\]\[\frac{-b \pm \sqrt{b^2-4a(c-y)}}{2a}=x\]Now we need to take the average value of x:\[\left(\frac{-b+\sqrt{b^2-4a(c-y)}}{2a}+\frac{-b-\sqrt{b^2-4a(c-y)}}{2a}\right)\frac{1}{2}=x_{v}\]\[x_{v}=\frac{-2b}{2a}\frac{1}{2}=-\frac{b}{2a}\]