complete the square: x^2+4x

To complete the square you must first make sure the coefficient of x^2 is 1, and if it is, then all you have to do is take 1/2 of the coefficient of the x term, in this case 4, and after taking 1/2 of 4, you square that result. This result is then added to the expression creating your perfect square.

\[x^2+ 4x+( \frac{4}{2})^2=x^2+ 4x+ 4\]

Exactly as Luis has demonstrated.

it has the answer set up like: (x+___)^2 + (___) ?

ok, so factor the result that Luis has, \[x ^{2}+4x+4=(x+2)^{2}+(-4)\]

You added 4 to the equation so you must subtract 4 from the erquation to keep it balanced.

equation*

thank you so much @calmat01 ! how would I graph that?

Ok, so your equation now looks like this \[y=(x+2)^{2}-4\] This is a parabola whose vertex occurs at the values x=-2 y=-4.

x=-2 comes from replacing x with -2 in the equation to make the expression in the parenthesis zero, and if what is inside is zero, then y must be -4.

Now once you have the vertex, just replace the x with two more values, one to the left of x=-2 and one to the right of x=-2.;

well it would be shifted left 2 units right and then down 4 units

That is correct. It's vertex is (-2, -4) so, just pick one more value to the left of x=-2 and one more value to the right of x=-2.

so would the parabola be facing up or down?

Can you answer that question based on our equation?

up because it's positive right

Exactly, the coefficient of the squared term is positive, so the graph points up. Good!

thank you so much for all of your help:)

No Prob. Glad I could help!

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