How do you graph the parabola #y= -3x^2 + 12x - 8# using vertex, intercepts and additional points?

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Answer 1 · 2018-02-15T09:24:00

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Every polinomic expresion of second degree is a parabola with general expresion #y=ax^2+bx+c#

The interception point with #y# axis we can obtain it appliying the value #x=0#. We obtain #y=-8#. So, our parabola inercepts
to #y# axis at #(0,-8)#

The intercetions with #x# axis we can obtain them appliying the value #y=0#. We get

#0=-3x^2+12x-8# we will obtain two points wich are the roots of that polinomical equation. In our case

#x=(-b+-sqrt(b^2-4ac))/(2a)=(-12+-sqrt(144+96))/-6#

#x=2-2/3sqrt15# and #x=2+2/3sqrt15#

So the intercets points with #x# axis are #(2-2/3sqrt15,0)# and #(2+2/3sqrt15,0)#

The vertex is given by #x=-b/(2a)=-12/-6=2# in wich value #y# reaches the value #y=4#

See the graph and zoom it (with mouse wheel) and pass the mouse cursor by intercept points, You will see the values we already get

graph{-3x^2+12x-8 [-22.99, 22.64, -16.14, 6.66]}