Question

put this into vertex form: y=3(x+2)(x-1)

Answer 1

\[y=3(x^2-x+2x-2)=3(x^2+x-2)=3(x^2+x)+3(-2)\]

Answer 2

how are you with this so far?

Answer 3

good. i put it into standard form so its \[y=3x ^{2}+3x-6\] but where do I go from there?

Answer 4

\[y=3(x^2+x)-6\]
this is what i have after multiplying 3 and -2

Answer 5

but is that vertex form?

Answer 6

\[y=3(x^2+1x+(\frac{1}{2})^2)-6-3(\frac{1}{2})^2\]
we wanted to complete the square inside that parenthesis so i added
(b/2)^2
but this was being multiplied by that 3
since I added 3(b/2)^2 I have to also subract 3(b/2)^2

Answer 7

no we have more work

Answer 8

\[y=a(x-h)^2+k <-\text{ this is vertex form }\]

Answer 9

\[y=3(x+\frac{1}{2})^2-6-3(\frac{1}{4})\]

Answer 10

\[y=3(x+\frac{1}{2})^2-6-\frac{3}{4}\]
\[y=3(x+\frac{1}{2})^2-\frac{24}{4}-\frac{3}{4}\]

Answer 11

\[y=3(x+\frac{1}{2})^2-\frac{27}{4}\]

Answer 12

where (h,k) is vertex

Answer 13

ok I see now :)

Answer 14

so the vertex is (-1/2, -27/4)?

Answer 15

yep thats right! :)

Answer 16

THANK YOU SO MUCH I FINALLY UNDERSTAND THIS