Question 375879
{{{y=3x^2+9x}}}
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Y=intercept:
{{{x=0}}
{{{y=3(0)+9(0)=0}}}
({{{0}}},{{{0}}})
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X-intercepts:
{{{3x^2+9x=0}}}
{{{3x(x+3)=0}}}
Two solutions:
{{{x=0}}}
({{{0}}},{{{0}}})
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{{{x+3=0}}}
{{{x=-3}}}
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({{{-3}}},{{{0}}})
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Convert to vertex form, {{{y=a(x-h)^2+k}}} by completing the square.
{{{y=3x^2+9x}}}
{{{y=3(x^2+3x)}}}
{{{y=3(x^2+3x+9/4)-3(9/4)}}}
{{{y=3(x+3/2)^2-27/4}}}
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Since {{{a=3>0}}}, the parabola opens upwards and the value at the vertex is the function minimum.
Range: ({{{-27/4}}},{{{infinity}}})
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{{{drawing(300,300,-8,8,-8,8,circle(0,0,0.3),circle(-3,0,0.3),circle(-3/2,-27/4,0.3),grid(1),graph(300,300,-8,8,-8,8,3x^2+9x,3(x+3/2)^2-27/4))}}}