#"to find the "color(blue)"intercepts"#
#• " let x = 0, in the equation for y-intercept"#
#• " let y = 0, in the equation for x-intercepts"#
#x=0tof(0)=9/2larrcolor(red)" y-intercept"#
#y=0to1/2x^2+3x+9/2=0#
#rArr1/2(x^2+6x+9)=0#
#rArr1/2(x+3)^2=0#
#x+3=0rArrx=-3larrcolor(red)" x-intercept"#
#"now "(x+3)^2=0rArrx=-3" of multiplicity 2"#
#"tells us that the graph has a turning point at "x=-3#
#rArrcolor(magenta)"vertex "=(-3,0)#
#color(blue)"shape of graph"#
#• " if "a>0" then minimum turning point "uuu#
#• " if "a<0" then maximum turning point "nnn#
# " here "x=1/2>0rArruuu#
#"choosing values of x and evaluating corresponding values"#
#"for y will give other points on the graph"#
#x=-1tof(-1)=1/2-3+9/2=2rArr(-1,2)" etc"#
graph{1/2x^2+3x+9/2 [-10, 10, -5, 5]}
