Question

Help convert to vertex/general form
f(x) = 2x^2 - 12x + 13 PLEASE

Answer 1

do you know what a "perfect square trinomial" is?

Answer 2

Yes

Answer 3

\(\bf f(x) = 2x^2 - 12x + 13\qquad \textit{let us begin by grouping the "x"'s}\\ \quad \\
f(x) = (2x^2 - 12x) + 13\implies f(x) = 2(x^2 - 6x) + 13\\\quad \textit{let's add a number}\\ \quad \\
f(x) = 2(x^2 - 6x+\color{red}{ \square}^2 ) + 13\)

so... what number there would give us a perfect square trinomial?

Answer 4

9 ?

Answer 5

well, yes, that is \(\bf 3^2\), yes, so
now keep in mind that all we're doing is borrowing from "0" zero, so if we add 9, we subtract 9, so 9-9=0

ab+c => ab+c+0 => ab+c+9-9 => ab+c

any expression if you add 0, is itself...so

\(\bf f(x) = 2x^2 - 12x + 13\qquad \textit{let us begin by grouping the "x"'s}\\ \quad \\
f(x) = (2x^2 - 12x) + 13\implies f(x) = 2(x^2 - 6x) + 13\quad \textit{let's add a number}\\ \quad \\
f(x) = 2(x^2 - 6x+3^2 ) + 13-3^2\implies f(x) = 2(x-3)^2+4\)

Answer 6

So, if you were to start by subtracting the 13 in the beginning, would the answer come out wrong?

Answer 7

well... if you subtract 13 at the beginning, you'd change the expression, and the equation would be changed, thus you'd lose the original equation

Answer 8

I keep getting the wrong answer by subtracting the 13 in the beginning, but when I look at my lesson, it says to subtract the 13 to isolate all x terms, then I have to complete the square, but I get it wrong.

Answer 9

well, subtracting the 13, or constant value, means \(\bf y = 2x^2 - 12x + 13\implies y-13 = 2x^2 - 12x\) which all it does is, let's you work with the "x"s, which you did anyway but just grouping, same thing though

so by "isolate 13" or the constant, you're really just isolating the "x"'s
which can also be done by grouping

Answer 10

After you plugged in the 3^2, how did you get four on the other side, I don't know how to get the four.

Answer 11

well just a correction as you removed a common factor of 2 before completing the square it needs to be

\[y = 2(x - 3)^2 + 13 - 2\times 3^2\]

otherwise you don't get the original equation... or the correct vertex

Answer 12

I don't know how to get the 4, as jdoee0001 got in his answer, and his answer is the same one I have to get on my practice test. Do you know how to get the four at the end @campbell_st

Answer 13

ok... so Jdoe's calculations have an error

so if you group the terms in x from your question you get

\[y = (2x^2 - 12x) + 13\]

so remove the common factor of 2 from the terms in x

\[y = 2(x^2 - 6x) + 13\]

you correctly said that 9 needs to be added to complete the square

so you have

\[y = 2(x^2 - 6x + 9) + 13\]

so in expanded form its

\[y = (2x^2 - 12x + 18)+ 13\]

I hope that makes sense so far.

Answer 14

Yes, I makes sense so far

Answer 15

it*

Answer 16

ok... to compensate adding 18 to the equation you need to subtract 18

so you have

\[y = 2x^2 - 12x + 18 + 13 - 18\]

now factoring you get

\[y = 2(x -3)^2 + 13 - 18\]

simplifying 13 - 18

and you get

\[y = 2(x -3)^2 - 5\]

which is different to the answer you were given, because of the error in only subtracting 9

hope it continues to make sense.

Answer 17

Yes :) Thank you , this helped me alot.