Question 364919
You're missing your exponent that makes it a quadratic.
I think this is what you mean.
{{{f(x)=3(x-2)^2-5}}}
SInce it's in vertex form and the coefficient of the {{{x^2}}} term is positive, the parabola opens upwards.
The value at the vertex, {{{-5}}}, is the function minimum.
Range:({{{-5}}},{{{infinity}}})
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{{{graph(300,300,-5,5,-8,2,0,3(x-2)^2-5)}}}