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General Form You have already seen how to graph a quadratic function when it is in its standard form. Recall that the standard form of a quadratic function looks like: y=a(x-h)^2+k But how do you graph a quadratic function when it is in its general form? The general form of a quadratic function looks like this: y=ax^2+bx+c
So how does your author define general form?
h=x, k=y... the problem I'm having with this one is splitting the 3 in 3x^2...?
Its just asking to rewrite it not graph it... heres what I have so far:
So you want h-k form?
-10 on both sides Y-10=3x^2-24x 24/2=12 12^2=144 Y-10=3x^2-24x+144
Its in standard form, I want general form(:
Let's write it like this: |dw:1362857059987:dw|
Now complete the square inside the parentheses
OH! gcf! of course!!:D
Thank you(: I forgot to look for a gcf
Y-10=3x^2-24x+144 Now factor the GCF y-10=3(x^2-8x+48) y-10=3(x-4)(x+12) +10 to both sides Y=3(x-4)(x+12) +10 Right?
how did you get 16^^
Oh wait... I know
No. You cannot add 48 to one side without adding it to the other side. Did you see in my example that I added 16 inside the parentheses to complete the square? But I was really adding 48 to the right side because there is a 3 outside the parentheses. So then I subtracted 48 so there would be a net change of 0. Of course I could have added 48 to the left side which would have accomplished the same purpose.
16 is the number that completes the square for x^2+8x
Y-10=3x^2-24x GCF Y-10=3(x^2-8x) Complete the square 8/2=4 4^2=16 y-10=3(x^2-8x-16) Simplify y-10=3(x-4)^2 +10 to both sides y=3(x-4)^2 +10 Right?
and by simplify I mean find the binomial...(:
So let's multiply out your result: \[y=3(x-4)^2+10=3(x^2-8x+16)+10=3x^2-24x+48+10=\]
Is that the problem you started with?
If it isn't, then what you have done is changed the original problem instead of keeping the original problem and expressing it in another form.
What am I doing wrong?
When you add 48 to the right side, you then have 2 choices: 1. subtract 48 from the right side 2. Add 48 to the left side.
y-10=3x^2-24x y-10=3x^2-24x+48 +48 y+38=3x^2-24x+48
y+38=3(x^2-8x+16) y+38=3(x-4)^2 Right so far?
Ok I solved it out on paper(: thank you for helping me
That is true. Not sure it is very useful.
k, never mind