Question

how do i write y=x^2+8x+15 in vertex form? please explain

Answer 1

do you know what a "perfect square trinomial" or sometimes called a "perfect square" is?

Answer 2

yes

Answer 3

ok... then let's start by grouping \(\bf y=x^2+8x+15\implies y=(x^2+8x)+15\implies y=(x^2+8x+{\color{red}{ \square }}^2)+15\) do you know what number we're missing to get a perfect square trinomial in the group?

Answer 4

no

Answer 5

well... recall that 8 = 2 * 2 * 2 => 2 * 4 thus \(\bf y=x^2+8x+15\implies y=(x^2+8x)+15\implies y=(x^2+8x+{\color{red}{ \square }}^2)+15
\\ \quad \\
y=(x^2+2(4)(x)+{\color{red}{ \square }}^2)+15\) what about now?

Answer 6

4?

Answer 7

yeap thus
now we know the 2nd term is "4" since the middle term of the perfect square is 2 * the other terms

so keep in mind that all we're doing is borrowing from our friend zero, 0,
so we'll ADD \(4^2\) BUT we also have to SUBTRACT \(4^2\)

thus \(\bf y=x^2+8x+15\implies y=(x^2+8x)+15\implies y=(x^2+8x+{\color{red}{ \square }}^2)+15
\\ \quad \\
y=(x^2+2(4)(x)+{\color{red}{ 4 }}^2)+15-{\color{red}{ 4 }}^2
\\ \quad \\
y=(x+4)^2+15-16\implies y=(x+4)^2-1\impliedby vertex\ form\)

Answer 8

thank you!

Answer 9

yw