Question

How do i write f(x) = x^2 + 6x + 1
in the form
f(x)= a(x - h)^2 + k

Answer 1

\(f(x) = \underbrace{x^2 + 6x} + 1\)

Complete the square by adding and subtracting a number that creates a perfect square binomial: \((x - a)^2\)

Answer 2

its not already a perfect square binomial?

Answer 3

We want to add to the \(x^2 + 6x\) something that makes it factor into \(\color{green}{(x + k)^2}\).

If we expand that out, we see \((x-k)^2 = x^2 + 2k x + k^2\).
We compare this to \(x^2 + 6x\), so 2k = 6; k = 1/2 of 6. Then squared gives us 9.

\(x^2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)^2\)

Sorry, mixed up the sign. We have + 6x, so it should be (x + k)

Answer 4

wait . how does that go into that equation a(x-h)^2+k

Answer 5

Once we know what to add to make the perfect square, we just add / subtract it:
\[ f(x) = x^2 + 6x + 1 \color{green}{+ 9 - 9} \]
Since adding / subtracting the same thing is still 0, it works out. Then, we just group the x^2 + 6x + 9 up and change it into (x + 3)^2.

\[ f(x) = (x^2 + 6x + 9) - 9 - 1 \\
f(x) = (x + 3)^2 - 10 \]

Answer 6

ohh. ok. i think i get that.

Answer 7

thank you

Answer 8

To maybe make more sense of things, basically, we look at the expression "x^2 + 6x," and add to it what would make it factor into something like "(x + #)^2. or "(x - #)^2" Usually, we just have to take half the coefficient on the "x" and square it to get the number to add.

Answer 9

Sorry if its a little hard to understand. This kind of thing is harder to explain than to just 'do'. :P

Answer 10

ahh, ok i remember hearing something like that though in the class. your right ha

Answer 11

thanks agn

Answer 12

You're welcome. :)

Answer 13

if youre still there, ha, so is there no a value in this case

Answer 14

Yeah, in this case, the a-value is just 1.
If we do have a coefficient on x^2, we have to factor it out like this:

\( \large f(x) = a(x^2 + \frac{b}{a}x + \frac{c}{a}) \)

The a-value is always the same as the coefficient on x^2.

Answer 15

ohh ok. i see. very helpful. a value always equal to the coefficient of x^2.. i can remember that.. appreciate it

Answer 16

Yeah. Just be careful when there is a coefficient other than 1, since you have to add/subtract inside the parentheses and then distribute the 'a' at the end.

(Later on, if you do conic sections, you do more completing the square stuff and have to add stuff outside of the parentheses, which is a bit more tricky since you're technically adding the constant term times 'a')

Answer 17

understood :)