Question

I'll fan and medal
Write the quadratic function in vertex form.

y = x2 - 16x + 66

Answer 1

@TheSmartOne @agent0smith

Answer 2

\[\Large y = x^2 - 16x + 66\]
is the same as
\[\Large y = 1x^2 +(-16x) + 66\]

Answer 3

\[\Large {\color{red}{1}}x^2+({\color{green}{-16}}x)+{\color{blue}{66}}\]
is in the form
\[\Large {\color{red}{a}}x^2+{\color{green}{b}}x+{\color{blue}{c}}\]

where
\[\Large \color{red}{a = 1}\]
\[\Large \color{green}{b = -16}\]
\[\Large \color{blue}{c = 66}\]

Answer 4

do you follow so far?

Answer 5

Yes @jim_thompson5910

Answer 6

What we do from here is plug the values of 'a' and 'b' into
\[\Large h = \frac{-b}{2a}\]
So,
\[\Large h = \frac{-(-16)}{2*1}\]
\[\Large h = ???\]
Tell me what you get

Answer 7

8

Answer 8

good

Answer 9

this means the x coordinate of the vertex is x = 8

plug x = 8 back into the original equation y = x^2 - 16x + 66 to find the corresponding value of y. Tell me what you get.

Answer 10

2

Answer 11

good

Answer 12

when x = 8, the corresponding paired value of y is y = 2
the vertex is therefore (h,k) = (8,2) which means h = 8, k = 2

now we know the following
a = 1
h = 8
k = 2
which we then plug into the vertex form template
\[\Large y = a(x-h)^2 + k\]

Answer 13

I got x^2-16x +66

Answer 14

So,
\[\Large y = a(x-h)^2 + k\]
turns into
\[\Large y = 1(x-8)^2 + 2\]
which simplifies to
\[\Large y = (x-8)^2 + 2\]

Answer 15

Don't expand out and combine like terms. Otherwise, you'll just get back to the original equation again.

Answer 16

Oh..Okay

Answer 17

The final answer is
\[\Large y = (x-8)^2 + 2\]

Answer 18

Thank you for helping, can you help with another?

Answer 19

Sure I can help. Please close this current question and start a new thread. It's best to have one question per thread.