The parabola \(y = ax^2 + bx + c\) is graphed below. Find \(a \cdot b \cdot c\). (The grid lines are one unit apart.)
0.1) { TicksArrx.push(i);} } for(i=ybottom+ystep; i 0.1) { TicksArry.push(i);} } if(usegrid) { xaxis(BottomTop(extend=false), Ticks("%", TicksArrx ,pTick=gray(0.22),extend=true),p=invisible);//,above=true);yaxis(LeftRight(extend=false),Ticks("%", TicksArry ,pTick=gray(0.22),extend=true), p=invisible);//,Arrows);} if(useticks) { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, Ticks("%",TicksArry , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));yequals(0, xmin=xleft, xmax=xright, p=axispen, Ticks("%",TicksArrx , pTick=black+0.8bp,Size=ticklength), above=true, Arrows(size=axisarrowsize));} else { xequals(0, ymin=ybottom, ymax=ytop, p=axispen, above=true, Arrows(size=axisarrowsize));yequals(0, xmin=xleft, xmax=xright, p=axispen, above=true, Arrows(size=axisarrowsize));} };rr_cartesian_axes(-7,1,-3,7);real f(real x) {return 1/2*x^2 + x + 3/2;} draw(shift((-2,-3))*graph(f,-5,3,operator ..), red);[/asy]" src="/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F3%2Fd%2Fc%2F3dcbfd37344accf75039423aaee0e135a89dc16a.png">
+37171 There are two 'zeroes' at x = -1 and -5 these are the roots
so the equation becomes:
(x+1)(x+5) = 0
x^2 +6x +5 = 0 y = x^2+6x+5 has the same roots , but not the same vertex
the vertex needs to be shifted up
the y value needs to be -2 at x = -3
(-3)^2 + 6(-3) + 5 = -4 Needs to be -2 multiply all of the coefficients by 1/2
1/2x^2 + 3x + 2.5 then when x = -3
4.5 + -9 + 2.5 = -2
y = 1/2x^2 + 3x + 2.5
+37171 You're kidding right????????????
y = a x^2 +bx + c
y = 1/2x^2 + 3x + 2.5
