how do you write y = 2x2 + 6x + 4 in general form?

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how do you write y = 2x2 + 6x + 4 in general form?

Mathematics
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do u know the general form?
yes, y=a(x-h)^2+k
did u need it in linear form?

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Other answers:

like the general form of a linear equation?
just the general form of a quadratic equation
okay, so u need to know the quadratic formula i believe..do u know it?
general form> y=a(x-h)^2+k
oh, yeah!
quadratic formula> -b+-sqrt(b-4(a)(c))/2(a)
but how does that help convert it to general form?
I thought it just helped find the x-intercepts..
if i can remember this correctly, i believe u need to first solve to get the values of a,b,c and then you just plug it into the general form... I'm not 100% sure tho... @abb0t, am i remembering this correctly? ;/
I know how to convert it... the problem is the 2 in 2x^2
General form of what?
quadratic equations
2 in 2x^2 is the problem.... you isolate the x variables getting: y-4=2x^2+6x then complete the square and balance the equation: 6/2=3 3^2=9, y-4+9=2x^2+6x+9, y+5=2x^2+6x+9 then convert the trinomial into a binomial... thus lies my dilemma
sorry, i'm not very sure, cuz i don't think i remember this correctly ahha and i don't wanna help u wrongly haha @Mertsj probs knows :) good luck!!! :D
thanks (:
\[y=2x^2+6x+4\] \[y=2(x^2+3x+______ )+4\]
Now complete the square by adding (3/2)^2 inside the parentheses: \[y=2(x^2+3x+(\frac{3}{2})^2)+4-\frac{9}{2}\]
Now factor: \[y=2(x+\frac{3}{2})^2-\frac{1}{2}\]
why 3/2?
I want to show you something: \[x^2+6x+9=(x+3)^2\]
\[x^2+8x+16=(x+4)^2\]
I don't get what you did with: (3/2^2)+4−92
Notice the relationship between the coefficient of x and the constant term in a trinomial square.
If you take 1/2 the coefficient and square it, you get the constant term.
So I took 1/2 of 3 and got 3/2. Then I squared it and added it to the x^2 + 3x to get a trinomial square.
Now the (3/2)^2 was inside a parenthesis which has a 2 in front of it so I was really adding 2 times (3/2)^2 which is 2 times 9/4 which is 9/2. So since I could not change the equation, I then had to subtract 9/2.
Follow me?
I think so
Okay, yes! I get what you did (: thank you! what were you trying to show me with those other two equations?
The relationship between the coefficient of x and the constant term.
oh, that half of the coefficient of x to the second power = the constant. okay, cool
Thank you for helping me (:
yw

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