How do you find the vertex of #y=x^2+9x+8#?

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Answer 1 · 2016-02-24T17:14:46

#color(blue)("A neat trick to find "x_("vertex"))#

Consider the standard for#" "y=ax^2+bx+c#.........(1)

rewritten as:#" "y=a(x^2+color(red)(b/ax))+c" ".............(1_a)#

Now compare it to your question's format

#" "y=x^2+color(green)(9x)+8#.......................(2)

In equation (2) #color(red)(a=1)# so for this equation we have

# color(red)(b/a x) -> color(green)((9x)/color(red)(1) = 9x#
'~~~~~~~~~~~~~~~~~~~~~~~~~
Hears the sneaky bit!

#x_("vertex")" "=" "(-1/2)xxa/b" " ->" " (-1/2)xx9 " "=""- 4 1/2#

So by substituting #x=-4 1/2# into equation (2) you can determine the value of #y#