Question

How do I find the standard form of this quadratic equation?

f(x) = 1/3 (x-3) (x+7)

Answer 1

multiply the right-side and then solve for "0"

Answer 2

http://www.mathsisfun.com/algebra/standard-form.html

Answer 3

I'm not sure what you mean.. Like, (x-3)(x+7)?

Answer 4

I'm reaaaaally bad at this stuff, sorry. :/

Answer 5

So like x^2 - 3x + 7x?

Answer 6

\(\bf f(x)=y=\frac{1}{3}(x-3)(x+7)\implies 0=\left[\frac{1}{3}(x-3)(x+7)\right]-y\)

Answer 7

is that the answer? I'm lost.. gahhh! I hate math >.<

Answer 8

though I'm thinking you're being asked for only to set the equation to 0... so that'd be \(\bf (x)=\frac{1}{3}(x-3)(x+7)\implies 0=\frac{1}{3}(x-3)(x+7)\)

Answer 9

and then you just expand the right side

Answer 10

What do you mean expand the right side? Like making 0 = 1/3 (x-3)(x+7) change to
0 = 1/3 x^2 - 3x + 7x?

Answer 11

Sorry I'm so stupid >.< I'm really bad a math but I reaally reaally appreciate the help love !<3 @jdoe0001

Answer 12

well... if you use FOIL so \(\bf f(x)=\frac{1}{3}(x-3)(x+7)\implies 0=\frac{1}{3}{\color{brown}{ (x-3)(x+7)}}
\\ \quad \\
{\color{brown}{ (x-3)(x+7)\to x^2+7x-3x-21\to x^2+4x-21}}
\\ \quad \\
0=\frac{1}{3}(x-3)(x+7)\implies 0=\frac{1}{3}(x^2+4x-21)\\ \quad \\\implies 0=\cfrac{(x^2+4x-21)}{3}\)

Answer 13

@jdoe0001 So the standard form equation would be: y= 1/3 x^2 + 4x -21 ?

Thanks so much! :)

Answer 14

yw