# Question #f4ccf

##### 2 Answers

Factored form:

Vertex form:

#### Explanation:

Yes, if the equation is equal to

#f(x) = x^2 -x - 42#

**Factored form** = gives you the zeros of a parabola.

Knowing that there is no

After trial and error, I got

#f(x)=x^2-7x+6x-42#

Now we use common factoring.

#f(x)=x(x-7)+6(x-7)#

We get the same number in the brackets - that's good. One bracket is one of the zeros, the coefficients are the other.

#f(x)=(x+6)(x-7)#

The factored form of

You can expand the equation to see if you get the original equation in standard form as a way to double check.

**Vertex form** = gives you the vertex of a parabola.

There are two ways to get the vertex form. One is to complete the square given an equation in standard form, the other is to find the axis of symmetry in factored form, and subbing in the value as

#f(x) = x^2 -x - 42#

First off, we factor out the

#f(x) = (x^2 -x) - 42#

Now, we divide the

#f(x) = (x^2 -x +0.25 - 0.25) - 42#

We expand the

#f(x) = (x^2 -x +0.25) -0.25 - 42#

We add like terms and we have our vertex form - the bracket is squared: so we'll "unsquare" it.

#f(x) = (x-0.5)^2 -42.25#

The vertex form of

You can expand the equation to see if you get the original equation in standard form as a way to double check.

Hope this helps :)

#### Explanation:

Factored gives

Vertex form

Use completing of the square:

Get the squared value by halving the x-term:

You need to subtract

From