Question 134725
Your function is in the form {{{f(x)=ax^2+bx+c}}} where {{{a=1}}}, {{{b=-1}}}, and {{{c=-12}}}


The vertex is located at the point ({{{(-b)/2a}}},{{{f((-b)/2a)}}}).  So, calculate {{{v[x]=(-b)/2a}}} to get the x-coordinate, then evaluate {{{f(v[x])}}} to get the y-coordinate.  Plot this point on your graph.


The x-intercepts are at (p,0) and (q,0) where {{{x=p}}} and {{{x=q}}} are the solutions to f(x)=0.  So solve {{{ x^2 - x - 12 = 0}}} to determine p and q, then plot these two points.


The y-intercept is at (0,f(0)).  Evaluate f(0) by substituting 0 for x in {{{x^2-x-12}}}.  Plot this point.


Multiply the x-coordinate of the vertex by 2.  Symmetry of the curve about the vertical line {{{x = (-b)/2a}}} means that there will be another point on the curve at ({{{(-b)/a}}},{{{f(0)}}}), i.e. 2 times the x-coordinate of the vertex and the y-coordinate of the y-intercept.  Plot this point.


Draw a smooth curve through the plotted points.  Done.


It should look something like this:


{{{drawing(600,600,-14,14,-14,14,
grid(1),
graph(600,600,-14,14,-14,14,x^2-x-12),
circle(4,0,.2),
circle(-3,0,.2),
circle(1/2,-49/4,.2),
circle(0,-12,.2),
circle(1,-12,.2)
)}}}