Solve #x^2-x-2=0# graphically, by first sketching #y=x^2#?

#color(blue)"sketching " y=x^2#

#"given a quadratic in standard form " ax^2+bx+c#

#• " if " a>0" then graph is a minimum " uuu#

#• " if " a<0" then graph is a maximum " nnn#

#"the coordinates of the vertex are " (-b/(2a),f(-b/(2a)))#

#"the y-intercept is the value of the constant c"#

#"the x-intercepts are found by equating to zero"#

#"for " y=x^2#

#a>0rArruuu#

#"the coordinates of the vertex are " (0,0)#

#"the value of " c=0#

#"some points on the graph "#

#x=+-1toy=1rArr(1,1),(-1,1)#

#x=+-2toy=4rArr(2,4),(-2,4)" etc"#
graph{x^2 [-10, 10, -5, 5]}

#color(blue)"sketching " y=x^2-x-2#

#• " for coefficient of x " >0#

#"the vertex moves " -b/(2a)larr" to the left"#

#• " for coefficient of x " <0#

#"the vertex moves " -b/(2a)rarr" to the right"#

#y=x^2-x-2 " is the same shape as " y=x^2#

#"coefficient of x term is " -1#

#rArrx_(color(red)"vertex")=--1/(2)=1/2rarr#

#rArry_(color(red)"vertex")=(1/2)^2-1/2-2=-9/4#

#rArrcolor(magenta)"vertex "=(1/2,-9/4)#

#"the y-intercept is at " c=-2rArr(0,-2)#

#"we could find the solution to " x^2-x-2" algebraically"#

#"by solving " x^2-x-2=0#

#"However, from the graph the solutions are the values "#
#"of x where the graph crosses the x-axis"#

#"that is " x=-1" or " x=2#
graph{x^2-x-2 [-10, 10, -5, 5]}