What value of a gives the parabola a minimum value of 1 in y=ax2+4x+3

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ax²+4x+3=

a(x²+4/a+3/a)=

a(x²+4/a+4/a²-4/a²+3/a)=

a(x+2/a)²-4/a+3.

The minimum value of this expression is -4/a+3 when x=-2/a. So -4/a+3=1, 4/a=2, so a=2.

The quadratic becomes 2x²+4x+3.

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