Question 318058
y = (-4x^2) - 12x + 9
Complete the square and put the eqaution in vertex form, {{{y=a(x-h)^2+k}}}
where (h,k) is the vertex.
{{{y= -4x^2-12x+9 }}}
{{{y=-4(x^2+3x)+9}}}
{{{y=-4(x^2+3x+9/4)+9+4(9/4)}}}
{{{y=-4(x+3/2)^2+18}}}
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a)The vertex is ({{{-3/2}}},{{{18}}}).
b) The parabola opens downward since the coefficient for {{{x^2}}} is negative.
c) Since the vertex occurs at y=18 and it opens downwards, the curve will cross the x-axis.
{{{y=-4(x+3/2)^2+18=0}}}
{{{4(x+3/2)^2=18}}}
{{{(x+3/2)^2=18/4}}}
{{{x+3/2=0 +- (3/2)sqrt(2)}}}
{{{highlight(x=-3/2 +- (3/2)sqrt(2))}}}
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{{{drawing(300,300,-5,2,-5,20,grid(1),
circle(0.62,0,.1),
circle(-3.62,0,.1),
circle(-1.5,18,.1),
graph(300,300,-5,2,-5,20, -4(x+3/2)^2+18,-4x^2-12x+9))}}}