SOLUTION: Hi, I know how to find the vertex I think, but I am not sure how to find the rest of the points to graph the equation. Can someone please help give some details about how to graph

Question 144376: Hi, I know how to find the vertex I think, but I am not sure how to find the rest of the points to graph the equation. Can someone please help give some details about how to graph this equation. I think the vertex is (-2,-8). Thanks.
f(x)= x^2+4x+4.
Thanks for your help.
Answer by nabla(475) (Show Source): You can put this solution on YOUR website!
Factor it first:
f(x)=(x+2)^2
The x-coordinate of the vertex will be where this function is valued least because the coefficient of x^2 is positive. That is, where (x+2)=0. That is x=-2. f(-2)=0. (-2,0) is the vertex.
Now, we know that x-intercepts are when the function is equal to 0. Thus we already found the only one: (-2,0)
For a y-intercept, we will set x=0. This gives f(0)=4. Therefore (0,4) is the y-intercept. This function will be symmetric about the line x=-2 (by definition). This is enough information to produce the following graph:

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