y=-(x-7)^2+10 is in vertex form, how do you put it in standard form?

Answer 1

Just multiply it out...

Start with the exponent:
(x-7)^2 = (x-7)(x-7) = x^2 - 14x + 49

Then you have the negative sign in front, so distribute that through:
-x^2 + 14x - 49

Finally you have the + 10 at the end. The final formula is:

y = -x^2 +14x - 39

Answer 2

Standard Form:

y=ax^2+bx+c

Here's how I would do it:

y= -(x-7)^2+10

y= -(x-7)(x-7)+10

Then use FOIL....

y= -(x^2-7x-7x+49)+10

y= -(x^2-14x+49)+10

y= -x^2+14x-49+10

y= -x^2+14x-39

There you have it...

Answer 3

y = -(x - 7)^2 + 10
y = -((x - 7)(x - 7)) + 10
y = -(x^2 - 14x + 49) + 10
y = -x^2 + 14x - 49 + 10
y = -x^2 + 14x - 39