Question 232780: f(x)= 7x^2+28x+28
i think i solved it this far.
f(x)=7(x^2+4x)+28
f(x)=7(x^2+4x+4-4)+ 7(-4)+28
f(x)=7(x^2+4x+4)+0
f(x)=7(x^2+4x)=
f(x)=7(x+2)^2
vertix is (-2,0)
so if i graph it x= -2 and y = 0
i have to use more points i tried
x= -1 y=7
x= -3 y= 7
x= -1.5 y= 1.75
i need to know if im rite and what other point i can use
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! >f(x)= 7x^2+28x+28
>i think i solved it this far.
>f(x)=7(x^2+4x)+28
>f(x)=7(x^2+4x+4-4)+ 7(-4)+28
>f(x)=7(x^2+4x+4)+0
>f(x)=7(x^2+4x)=
>f(x)=7(x+2)^2
This is actually correct. But I do not understand the steps you took to get here. I almost think you got the right equation by accident. Here's a short, easy way:
f(x)= 7x^2+28x+28
Factor out 7 from all the terms:
f(x) = 7(x^2 + 4x + 4)
The second factor fits the perfect square trinomial pattern: a^2 + 2ab + b^2. So now we have:
f(x) = 7(x+2)^2
vertex is (-2,0)
so if i graph it x= -2 and y = 0
i have to use more points i tried
x= -1 y=7
x= -3 y= 7
x= -1.5 y= 1.75
i need to know if im rite and what other point i can use
I recommend that you find matching points on either side of the axis of symmetry. Since x = -2 is the axis of symmetry, use -1 and -3 (1 to the right and 1 to the left of -2), 0 and -4 (2 to the right and to the left of -2), etc. Once you get the idea of parabolas, you'll find, for example, that f(0) = f(-4) so you only need to figure out one of them to know both of them.
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