Answer:
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.
