Question

How do I convert y=2x^2-4x+7 into vertex form?

Answer 1

Vertex form:
\[y = a(x - h)^{2} + k\]

Answer 2

You have your equation in standard form.

Answer 3

I think, it's in standard form right?

Answer 4

I forgot that equation.

Answer 5

Yes it is

Answer 6

Cool, let me think...

Answer 7

in standard form, x is the axis of symmetry, correct?

Answer 8

Yeah, it is. hmmz

Answer 9

\[x = -\frac{ b }{ 2a }\]
is the axis of symmetry.

Answer 10

\[y = 2x^{2} - 4x + 7\]
\[x = -\frac{ b }{ 2a }\]
\[x = -\frac{ -4 }{ 2(2) }\]
\[x = -\frac{ -4}{ 4 }\]

Answer 11

With me so far? Any questions?

Answer 12

Possibly but hold on

Answer 13

Whenever you're ready. :)

Answer 14

Can you please check this

Answer 15

One second, taking my dog outside.

Answer 16

Okay, checking.

Answer 17

Your work looks correct to me. :D

Answer 18

Ok thank you

Answer 19

No problem.

Answer 20

Y = 2x^2 - 4x + 7

I normally start by subtracting 7 from both sides, -7 = 2x^2 - 4x

Now you can factor this: 2x^2 - 4x -> 2(x^2 - 2x)

Then taking half the "x" term, and then squaring it. \(\ \sf -7 =2(x^2 - 2\color{blue}{x})\)

So, \(\ \dfrac{-2}{2} = -1, -1^2 = 1 \) therefore, I add 1 into the parenthesis,

-7 = 2([x^2 - 2x + 1] -1) ... you added 1 into the parenthesis, to keep it balanced you subtract 1 outside it.

Now factor x^2 - 2 + 1, you'll always get (x \(\ \pm \) ? )^2

Where the "?" is, you'll always have half the "x" term, so -2/2 = -1, so:

(x - 1)^2 is the factored form of x^2 - 2x + 1.

-7 = 2[(x - 1)^2 - 1], distribute the 2 to -1, 2*-1 = -2

-7 = 2(x - 1)^2 - 2

Add 7 to the RHS,

-7 = 2(x - 1)^2 - 2
+7 +7

y = 2(x - 1)^2 + 5

And that's vertex form, \(\ \Large y = a(x - h)^2 + k\)

The vertex is at: (h,k)
The axis of symmetry is the x coordinate in the vertex.

Vertex: (1, 5)
Axis: x = 1