Question

y=x^2+2x+1 find the vertex

Answer 1

2x + 2 = 0
x = -1

(-1, 0)

Answer 2

vertex is
\[(-1,0)\] because
\[y=x^2+2x+1=(x+1)^2\]

Answer 3

in general vertex is
\[x=-\frac{b}{2a}\] and then y is whatever you get when you replace x by
\[-\frac{b}{2a}\]

Answer 4

the axis of symmetry?

Answer 5

it is
\[x=-\frac{b}{2a}\] again. in this case it is
\[x=-1\]

Answer 6

would that be a minimum or maximum value? and what would the value be? and what is the range of the parabola

Answer 7

these are all essentially the same question, just written in different forms.
a) the vertex is (-1,0)
b) the axis of symmetry is x = -1
c) the minimum value is 0 (the second coordinate of a vertex, and it is the minimum because this parabola opens up
d) the range is
\[[0,\infty)\] for the same reason as answer c

Answer 8

thank you:)!

Answer 9

thank you:)!

Answer 10

yw