Question

how do I get y=3x^2-30x+69 into standard form y=a(x-h)^2+k ?????

i keep getting the answer y=3(x-5)^2 +44 and that's wrong....apparently its supposed to be y=3(x-5)^2-6

Answer 1

If we re-expanded \[3(x-5)^2+44 = 119 - 30 x + 3 x^2\]
So, \[3(x-5)^2=75 - 30 x + 3 x^2\] and the original equation had +69. Where can we go from here?

Answer 2

Or, e.g 75-??=69

Answer 3

i am trying to get y=3x^2-30x+69 into standard form y=a(x-h)^2+k

Answer 4

look at your quadratic equation and split it \[(3x^2 + 30x) + 69\]

next remove a factor of 3 from the brackets

\[3(x^2 + 10x + ??) + 69 - 3??\]

inside the brackets needs to be a perfect square... so you need to add a value, where ?? is inside the brackets

the same value needs to be subtracted outside the brackets to keep the equation in balance 3??

the reason iwhy its 3?? is because you have a factor of 3 outside the brackets.

Hope it helps.

Answer 5

oops should be

\[(3(x^2 - 10x + ??) + 69 - 3\times ??\]

Answer 6

The \[3(x−5)^2\] part is right. It's just the constant term that wasn't calculated correctly, \[75+44 \ne 69 \]

Answer 7

i know the 3(x-5)^2 part is right, but I dont know how i get the "k" term. k should equal -6

Answer 8

75+??=69, just ask yourself how to correct the extra you've added to get 3(x-5)^2

Answer 9

Or, \[3x^2+30x+69+6\] you can factor easily. But you can't change the equation. So\\[(3x^2+30x+69+6)-6\]

Answer 10

figured it out on khan academy: https://www.khanacademy.org/math/algebra/quadratics/features-of-quadratic-functions/v/ex3-completing-the-square

1. y=3x-30x+69

2. y=3(x-10x+23)

3. y=3(x-10x+25-25+23)

4. y=3(x-10x+25) 3(-25-23)

5. y=3(x-5)^2 3(-2)

6. y=3(x-5)^2 -6

Answer 11

if you complete the square you get

\[3(x^2 - 10x + 25) + 69\]

so the adjustment needed to compensate is

\[3(x^2 - 10x + 25) + 69 - 75\]

which becomes

\[3(x - 5)^2 - 6\]

Answer 12

remember that is you distribute

\[3(x^2 - 10x + 25) = 3x^2 - 30x + 75\]

so the compensation is - 75