Where is the vertex of the parabola?

y = x2 + 2x + 3

a. above the x-axis

b. on the x-axis

c. on the y-axis

d. below the x-axis

ABOVE the x-axis

Step-by-step explanation:

Please use "^" to denote exponentiation:  y = x^2 + 2x + 3

To find the vertex, we must complete the square of y = x^2 + 2x + 3, so that we have an equivalent equation in the form f(x) = (x - h)^2 + k.

Starting with  y = x^2 + 2x + 3,

we identify the coefficient of x (which is 2), take half of that (which gives

us 1),  add 1 and then subtract 1, between "2x" and "3":

y = x^2 + 2x + 1 - 1 + 3

Now rewrite x^2 + 2x + 1 as (x + 1)^2:

y = (x + 1)^2 - 1 + 3, or y = (x + 1)^2 + 2.  Comparing this to f(x) = (x - h)^2 + k, we see that h = 1 and k = 2.  This tells us that the vertex of this parabola is at (h, k):  (1, 2), which is ABOVE the x-axis.

i think it is c, not positive though

step-by-step explanation:

3; 9 is 8 to -4

step-by-step explanation: