## Algebra 1

a) 7.4 $units^2$ b) 7.8 $units^2$ c) $(1.61, 3.22)$, $(1.61, -3.22)$, $(-1.61, 3.22)$, $(-1.61, -3.22)$ d) 10.37 $units^2$
a) length of base is 2 units $y=-.3x^2+4$ $y=-.3(1)^2+4$ $y=-.3*1+4$ $y=-.3+4$ $y=3.7$ $3.7*2 = 7.4$ $units^2$ b) length of base is 6 units $y=-.3x^2+4$ $y=-.3(3)^2+4$ $y=-.3*9+4$ $y=-2.7+4$ $y=1.3$ $6*1.3 = 7.8$ $units^2$ c) Since we are looking for a square, we have to set two equations equal to each other. $y=2x$ $y=-.3x^2+4$ $2x=-.3x^2+4$ $.3x^2+2x-4=0$ $10*(.3x^2+2x-4=0)$ $3x^2+20x-40=0$ $x=(-b±\sqrt {b^2-4ac})/2a$ $x=(-20±\sqrt {20^2-4*3*-40})/2*3$ $x=(-20±\sqrt {400+480})/6$ $x=(-20±\sqrt {880})/6$ $x=(-20±29.66)/6$ $x=(-20+29.66)/6$ $x=9.66/6$ $x=1.61$ $x=(-20-29.66)/6$ $x=-49.66/6$ $x=-8.28$ (we can't have a negative length, so this answer is invalid) $x=1.61$ $y=2x$ $y=2*1.61$ $y=3.22$ $(1.61, 3.22)$ Since the square is symmetric, we can find the other three points to be $(1.61, -3.22)$, $(-1.61, 3.22)$, and $(-1.61, -3.22)$ d) $1.61-(-1.61) = 3.22$ $3.22*3.22 = 10.37$