# Question f4ccf

Mar 27, 2017

Factored form: $f \left(x\right) = \left(x + 6\right) \left(x - 7\right)$.
Vertex form: $f \left(x\right) = {\left(x - 0.5\right)}^{2} - 42.25$.

#### Explanation:

Yes, if the equation is equal to $y$ AKA f(x). Assuming it is...

$f \left(x\right) = {x}^{2} - x - 42$

Factored form = gives you the zeros of a parabola.

Knowing that there is no $a \text{-value}$, we can use a more simpler form of decomposition. Considering that this is a simple trinomial, we only need to do the first half of decomposition: what two numbers multiplied equals $a c$ and added equals $b$.

After trial and error, I got $- 7$ and $6$, both fit the requirement ^ .

$f \left(x\right) = {x}^{2} - 7 x + 6 x - 42$

Now we use common factoring.

$f \left(x\right) = x \left(x - 7\right) + 6 \left(x - 7\right)$

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

$f \left(x\right) = \left(x + 6\right) \left(x - 7\right)$

The factored form of $f \left(x\right) = {x}^{2} - x - 42$ is >$f \left(x\right) = \left(x + 6\right) \left(x - 7\right)$.

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 $x$ into the equation. We will complete the square.

$f \left(x\right) = {x}^{2} - x - 42$

First off, we factor out the $a \text{-value}$ from the standard form equation. Our $a \text{-value}$ is one, but no matter. It sets us up for the next step.

$f \left(x\right) = \left({x}^{2} - x\right) - 42$

Now, we divide the $b \text{-value}$ by two, and square it. Now we can't just add in a value, thus, we have to subtract the $0.25$ that we added.

$f \left(x\right) = \left({x}^{2} - x + 0.25 - 0.25\right) - 42$

We expand the $- 0.25$.

$f \left(x\right) = \left({x}^{2} - x + 0.25\right) - 0.25 - 42$

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

$f \left(x\right) = {\left(x - 0.5\right)}^{2} - 42.25$

The vertex form of $f \left(x\right) = {x}^{2} - x - 42$ is $f \left(x\right) = {\left(x - 0.5\right)}^{2} - 42.25$.

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 :)

Mar 27, 2017

$\left(x + 6\right) \left(x - 7\right) = 0$
$y = {\left(x - \frac{1}{2}\right)}^{2} - \frac{169}{4}$

#### Explanation:

Factored gives $x$-intercepts: $\left(x + 6\right) \left(x - 7\right) = 0$
$x$-intercepts: $\left(- 6 , 0\right) , \left(7 , 0\right)$

Vertex form $y = a {\left(x - h\right)}^{2} + k$, where vertex: $\left(h , k\right)$

Use completing of the square:
$\left({x}^{2} - x\right) - 42 = 0$

Get the squared value by halving the x-term: $\frac{1}{2} \cdot - 1 = - \frac{1}{2}$
${\left(x - \frac{1}{2}\right)}^{2} - 42 - \frac{1}{4} = 0$

You need to subtract ${\left(\frac{1}{2}\right)}^{2}$ because it is added when the square is completed: ${\left(x - \frac{1}{2}\right)}^{2} = \left(x - \frac{1}{2}\right) \left(x - \frac{1}{2}\right) = {x}^{2} - x + \frac{1}{4}$

From $\text{ } {\left(x - \frac{1}{2}\right)}^{2} - 42 - \frac{1}{4} = 0$
$y = {\left(x - \frac{1}{2}\right)}^{2} - \frac{169}{4}$