Which of the following tables shows the correct steps to transform x2 + 6x + 8 = 0 into the form (x - p)2 = q? [p and q are integers]
Step 1 x2 + 6x + 8 - 1 = 0 - 1 Step 2 x2 + 6x + 7 = -1 Step 3 (x + 3)2 = -1 Step 1 x2 + 6x + 8 + 1 = 0 + 1 Step 2 x2 + 6x + 9 = 1 Step 3 (x + 3)2 = 1 Step 1 x2 + 6x + 8 - 2 = 0 - 2 Step 2 x2 + 6x + 6 = -2 Step 3 (x + 3)2 = -2 Step 1 x2 + 6x + 8 + 2 = 0 + 2 Step 2 x2 + 6x + 10 = 2 Step 3 (x + 3)2 = 2
@amistre64 @ganeshie8 @beccaboo333
well, how would you go about it?
what is (x+3)^2?
what is it factored?
no, what is (x+3)^2 when we expand it into a trinomial? if FOIL sounds familar then work it
I haven't learned foil
have you learned to multiply binomials?
yes
all FOIL is is some memory device used to remember how to multiply binomials .... i find it to be utterly useless myself. (x+3)^2 = (x+3)(x+3) = ____ + _____ + _____
so multiply (x+3)(x+3)?
of course ....
we are simply working backwards to see what step2 needs to look like.
x^2+6x+9
correct, now which options has that as the correct step2?
i don't see one that does
wait nvm its the second one
:) yeah
okay thank you , do you think you could try and help with another question?
maybe one more
Functions f(x) and g(x) are shown below: f(x) = x2 g(x) = x2 - 10x + 25 In which direction and by how many units should f(x) be shifted to obtain g(x)? Right by 5 units Left by 5 units Right by 10 units Left by 10 units
I take that as a no?
take what as a no?
let x=0, what do we get?
for which one
both of them ... we can compare the results. unless you know of a different way to approach it.
do you know how to find a vertex yet?
the first one is 0 the second one is 15 and yes
its either the first or second one
lets determine the x value of the vertex for g then, that should help determine where they moved it to :)
(5,0) is the vertex for g
then they moved f(0,0) to g(5,0) which option shows this movement?
b?
|dw:1401228966945:dw| i think its moved to the right by 5 .... as long as my directions arent messsed up lol
gotcha, lol thank you!!
youre welcome ;)
@mathmorelikeno
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