Question 1058338
False, points on the parabola are solutions.
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{{{y=5(x^2+(6/5)x)+2}}}
{{{y=5(x^2+(6/5)x+(3/5)^2)+2-5(3/5)^2}}}
{{{y=5(x+3/5)^2+2-9/5}}}
{{{y=5(x+3/5)^2+1/5}}}
True, the vertex is ({{{-3/5}}},{{{1/5}}}).
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False, parabola opens upwards.
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{{{0<=5(0)^2+6(0)+2}}}
{{{0<=2}}}
It's true that 2 is greater than 0 so (0,0) is a solution.
False.