# What is the vertex of y=x^2-2x-35?

Feb 29, 2016

color(blue)("Vertex" -> (x,y)-> (1,-36)

I have shown a really 'cool' trick to solve this.

#### Explanation:

To demonstrate how useful the method I am about to show you is.

Just by looking at the given equation I determine that ${x}_{\text{vertex}}$ is at $x = + 1$

Then it is just a matter of substitution to find ${y}_{\text{vertex}}$ which is -36
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Ok! lets deal with the method!

The standardised equation structure $a {x}^{2} + b x + c$

In you case $a = 1$

Change $a {x}^{2} + b x + c \text{ to } a \left({x}^{2} + \frac{b}{a} x\right) + c$

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$
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$\textcolor{b l u e}{\text{Solving your question}}$

$a = 1$

$\frac{b}{a} = \frac{- 2}{1} = - 2$

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 2\right) = \textcolor{red}{+ 1}$

$\textcolor{b r o w n}{\text{The above is part way to developing the vertex equation format}}$

Sometimes the number are a bit more difficult to work out.

${y}_{\text{vertex}} = {\left(\textcolor{red}{1}\right)}^{2} - 2 \left(\textcolor{red}{1}\right) - 35$

${y}_{\text{vertex}} = 1 - 2 - 35 = - 36$

color(blue)("Vertex" -> (x,y)-> (1,-36)