Question 138797

find the vertex of the parabola represented by each equation. 
1) {{{y=2x^2-6x+6}}} 

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The vertex of the parabola whose equation is {{{y=Ax^2+Bx+C}}} is

({{{-B/(2A)}}}, {{{-D/(4A)}}})

where {{{D}}} = the discriminant {{{B^2-4AC}}}

For {{{y=2x^2-6x+6}}},

{{{A=2}}}, {{{B=-6}}}, {{{C=6}}} 

{{{D = B^2-4AC = (-6)^2-4(2)(6) = 36-48 = -12}}}

So vertex = ({{{-B/(2A)}}}, {{{ -D/(4A) }}} ) = ({{{-(-6)/(2(2))}}}, {{{-(-12)/(4(2))}}}) = ({{{6/4}}}, {{{12/8}}}) = ({{{3/2}}}, {{{3/2}}})
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2) {{{y=3x^2+4x=5}}}

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You mistyped that one because it should not have two equal signs. But it's 
done the same way.

If the second equal sign was supposed to have been a +, then 

vertex = ({{{-2/3}}},{{{11/3}}})

If the second equal sign was supposed to have been a -, then 

vertex = ({{{-2/3}}},{{{-19/3}}})

Edwin</pre>