How do you find the vertex of $g(x)= 5/7x^2+ 13/21x + 1/3$ ?

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Answer 1 · 2016-05-04T13:08:54

$color(blue)("Vertex "->(x,y)->(-13/30,+251/1260)" Exactly"$

$color(blue)("Vertex "->(x,y)->(-0.43bar3, 0.199)" Approximately"$

$color(blue)("Determine "x_("vertex"))$

Write as $" "y=5/7[x^2+(13/21-:5/7)x] + 1/3$

$y= 5/7(x^2+13/15x)+1/3$

$color(blue)(x_("vertex")=(-1/2)xx13/15 = -13/30)$

'.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.
$color(blue)("Determine "y_("vertex"))$

Substitute for $x$ in the original equation

$y_("vertex")=5/7(-13/30)^2+13/21(-13/30)+1/3$

$y_("vertex") = (5/7xx160/900)-(13/21xx13/30)+1/3$

$y_("vertex")=169/1260-169/630+1/3$

$color(blue)(y_("vertex")=251/1260 ~~0.1992)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("Vertex "->(x,y)->(-13/30,+251/1260)" Exactly"$
$color(red)("Thank goodness for calculators!")$

$color(blue)("Vertex "->(x,y)->(-0.43bar3, 0.199)" Approximately"$

Tony B

How do you find the vertex of g(x)= 5/7x^2+ 13/21x + 1/3g(x)= 5/7x^2+ 13/21x + 1/3?