Question

help . . .where is the vertex of y=1/2x^2+2x-3?

Answer 1

derivate this expression one time
y'=2/2x+2
the vertex is for y'=0
so x=-2

Answer 2

magic formula for vertex is
\[-\frac{b}{2a}\]

Answer 3

now solve
y=1/2 x^2+2x-3 with x=-2
y=-5

Answer 4

in this case
\[a=\frac{1}{2},b=2\] so
\[-\frac{b}{2a}=-\frac{2}{2\times \frac{1}{2}}=-2\]

Answer 5

first coordinate of vertex is -2, second is what you get when you replace x by -2 in the original expression
\[\frac{1}{2}\times (-2)^2+2\times -2-3\]
\[2-4-3\]
\[-5\] so vertex is
\[(-2,-5)\]

Answer 6

i suppose if you know calculus you can use it to find the vertex, but you certainly don't need it. just remember
\[-\frac{b}{2a}\]

Answer 7

thanks satellite!