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.

your answer is d 2,500

step-by-step explanation:

answer: on graphs, the open and closed circles, or vertical asymptotes drawn as dashed lines us identify discontinuities.

as before, graphs and tables allow us to estimate at best.

when working with formulas, getting zero in the denominator indicates a point of discontinuity.

when working with piecewise-defined functions, check for discontinuities at the transition points where one piece ends and the next begins.

step-by-step explanation: