Question 379796
When {{{x=0}}}, {{{y=42}}}
({{{0}}},{{{42}}})
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Complete the square to convert the equation to vertex form, {{{y=a(x-h)^2+k}}} in order to deterine the axis of symmetry.
{{{f(x)=10x^2+40x+42}}}
{{{f(x)=10(x^2+4x)+42}}}
{{{f(x)=10(x^2+4x+4)+42-10(4)}}}
{{{f(x)=10(x+2)^2+2}}}
The vertex is ({{{-2}}},{{{2}}}) and the vertex lies on the axis of symmetry {{{x=-2}}}.
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{{{drawing(300,300,-6,2,-2,18,grid(1),circle(-2,2,0.13),blue(line(-2,-20,-2,20)),graph(300,300,-6,2,-2,18,10x^2+40x+42))}}}