How do you convert y+8=2(x-1)^2 into standard form? Can someone show me step by step?

Question

ohmybookness · Answer

@phi can you help me out here?

ohmybookness · Answer

I know the coordinates of the vertex is (1,8)

phi · Answer

can you multiply out
(x-1)(x-1)
?

ohmybookness · Answer

x^2-2x+1

phi · Answer

now multiply each term by 2
(we are expanding 2(x-1)^2 )

phi · Answer

you should get
$$ 2x^2 -4x+2$$
so now you have
$$ y+8= 2x^2 -4x+2$$

the last step is add -8 to both sides

phi · Answer

btw, the vertex is at (1,-8)

ohmybookness · Answer

oops my bad

ohmybookness · Answer

@phi is it y=2(x-1)^2-8

jdoe0001 · Answer

note:  $\bf y+8=2(x-1)^2 \cong =2(x-1)^2-8$

phi · Answer

that is "vertex form"
standard form is ax^2+bx + c
which we almost have (see post up above)
y+8= 2x^2 -4x+2

jdoe0001 · Answer

$\bf y+8=2(x-1)^2 \cong y=2(x-1)^2-8$  that is

ohmybookness · Answer

wait so what i wrote was not standard form? huh. I thought it was

phi · Answer

phi · Answer

what was the original question?

ohmybookness · Answer

to convert y+8=2(x-1)^2 into standard form

phi · Answer

ok, then you are almost done. so far you have
y+8= 2x^2 -4x+2

ohmybookness · Answer

yeah

phi · Answer

add -8 to both sides

phi · Answer

y+8-8= 2x^2 -4x+2-8

now simplify to get "standard form"