What is the vertex form of y=(-x-1)(x+7)?

Dec 23, 2016

$\text{Vertex form "->" } y = - 1 {\left(x \textcolor{m a \ge n t a}{- 3}\right)}^{2} \textcolor{b l u e}{+ 2}$

$\text{Vertex} \to \left(x , y\right) = \left(3 , 2\right)$

Explanation:

First return this to the form of $y = a {x}^{2} + b x + c$

$y = \textcolor{b l u e}{\left(- x - 1\right)} \textcolor{b r o w n}{\left(x + 7\right)}$

Multiply everything in the right hand bracket by everything in the left.

$y = \textcolor{b r o w n}{\textcolor{b l u e}{- x} \left(x + 7\right) \textcolor{b l u e}{\text{ } - 1} \left(x + 7\right)}$

$y = - {x}^{2} + 7 x \text{ } - x - 7$

$y = - {x}^{2} + 6 x - 7. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$
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Write as: $y = - 1 \left({x}^{2} - 6 x\right) - 7 + k$
The $k$ corrects the error this process introduces.

Move the power from ${x}^{2}$ to the outside of the btackets

$y = - 1 {\left(x - 6 x\right)}^{2} - 7 + k$

Halve the 6 from $6 x$

$y = - 1 {\left(x - 3 x\right)}^{2} - 7 + k$

Remove the $x$ from the $3 x$

$y = - 1 {\left(x - 3\right)}^{2} - 7 + k \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left({1}_{a}\right)$
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Dealing with the error
If you were to expand the brackets and multiply by the -1 you have the value of $\left(- 1\right) {\left(- 3\right)}^{2} = - 9$. Looking back at $E q u a t i o n \left(1\right)$ you will observe that this value is not in it. So we have to remove the $- 9$

Set $- 9 + k = 0 \implies k = 9$
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Substitute for $k \text{ in } E q u a t i o n \left({1}_{a}\right)$

$y = - 1 {\left(x - 3\right)}^{2} - 7 + k \textcolor{g r e e n}{\text{ "->" } y = - 1 {\left(x - 3\right)}^{2} - 7 + 9}$

$y = - 1 {\left(x \textcolor{m a \ge n t a}{- 3}\right)}^{2} \textcolor{b l u e}{+ 2}$

${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{m a \ge n t a}{\left(- 3\right)} = + 3$
${y}_{\text{vertex}} = \textcolor{b l u e}{+ 2}$

$\text{Vertex} \to \left(x , y\right) = \left(3 , 2\right)$