Question

f(x) = −3x^2 − 18x − 29 put into a(x-h)^2+k form

Answer 1

you are supposed to complete the square to find the answer. so far i have -3(x2-6x+9)-29-9

Answer 2

i have to find the standard equation for a parabola with vertical axis

Answer 3

a(x-h)² +k =ax²-2axh+ah²+k=-3x²-18x-29
so a=-3
-2ah=-18
ah²+k=-29

Answer 4

the equality of two polynomes , it's a mathematic propriety ; you will find a , h and k

Answer 5

? that is not the way my book shows

Answer 6

factor out the \(-3\) get
\[(x)=-3(x^2+6x)-29\] the half of 6 is 3 and write
\[f(x)=-3(x+3)^2+k\]
find \(k\) by replacing \(x\) by \(-3\) in the original function

Answer 7

the problem with the method you wrote as a first step, is when you completed the square you thought you added a 9 and therefore subtracted a 9
but in fact you did not add 9 you subtracted 27 (distributive property) so you have to add 27

Answer 8

a=-3
h=-3
k=-2
by solving systema

Answer 9

you wrote
\[-3(x^2-6x+9)-29-9\] but the +9 is actually \(-27\) since it is being multiplied by \(-3\)

Answer 10

you should write
\[-3(x^2+6x+9)-29+27\]

Answer 11

-3(x+3)^2-27?

Answer 12

notice also that when you factor out the \(-3\) is not \(-6x\) but rather \(+6x\)

Answer 13

there is this method ; of equality of two polynomes

Answer 14

\(-29+27=-2\)

Answer 15

\[f(x)=-3(x+3)^2-2\] should do it
that is why i think it is easier to evaluate the function at \(-3\) then to try to keep track of the constant

Answer 16

thanks :)

Answer 17

you can avoid the common mistake of not realizing what you added or subtracted by just starting with
\[f(x)=-3(x+3)^2+k\] and since clearly
\[f(-3)=k\] you can find \(k\) by finding \(f(-3)\) using the original expression for \(f(x)\)