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 $y$

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