Graph the function g(x) =x^2+4x+1

Step-by-step explanation:

Let's "complete the square," which will give us the vertex of this vertical, opens-up parabola:

g(x) =x^2+4x+1 can be rewritten as g(x) =x^2+4x + 4 - 4 +1, where that +4 comes from squaring half of the coefficient of x.

Then we have   g(x) =x^2+4x + 4 - 4 + 1  =>  g(x) = (x + 2)^2 - 3.

Comparing this to                                             y   = (x - h)^2 + k,

we see that the vertex, (h, k), is located at (-2, -3).

Plot this vertex.  Also, plot the y-intercept (0, g(0) ), which is (0, 1).

This information is enough to permit graphing the function roughly.

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step-by-step explanation:

your welcome : )

step-by-step explanation:

do you have a copy of the table?