Question

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

Answer 1

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

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

Explanation:

`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"`