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

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Answer 1 · 2016-02-29T13:30:26

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

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

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

Just by looking at the given equation I determine that #x_("vertex")# is at #x=+1#

Then it is just a matter of substitution to find #y_("vertex")# which is -36
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Ok! lets deal with the method!

The standardised equation structure #ax^2+bx+c#

In you case #a=1#

Change #ax^2+bx+c" to " a(x^2+b/a x) +c#

Then #x_("vertex") = (-1/2)xx b/a#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving your question")#

#a=1#

#b/a = (-2)/1=-2#

#x_("vertex")=(-1/2)xx(-2) = color(red)(+1)#

#color(brown)("The above is part way to developing the vertex equation format")#

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

#y_("vertex")=(color(red)(1))^2-2(color(red)(1))-35#

#y_("vertex")=1-2-35=-36#

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