How do you solve #x^2 - x - 12 = 0# graphically?
1 Answer
Dec 23, 2017
Explanation:
#"given the graph of "x^2-x-12#
#"the solutions are the values of x where the graph crosses"#
#"the x-axis"#
graph{(y-x^2+x+12)((x+3)^2+(y-0)^2-0.04)((x-4)^2+(y-0)^2-0.04)=0 [-10, 10, -5, 5]}
#rArrx=-3" or "x=4" are the solutions"#
#"we can verify this algebraically by solving"#
#x^2-x-12=0#
#rArr(x-4)(x+3)=0#
#x-4=0rArrx=4#
#x+3=0rArrx=-3#
