#f(x) =-(x+2)(x-5)#

The first two 'important points' are the zeros of #f(x)#. These occur where #f(x)=0# - I.e. the #x-#intercepts of the function.

To find the zeros: #-(x+2)(x-5) =0#

#:.x=-2 or 5#

Hence the #x-#intercepts are: #(-2, 0) and (5, 0)#

Expanding #f(x)#

#f(x) = -x^2+3x+10#

#f(x)# is a quadratic function of the form #ax^2+bx+c#. Such a function is represented graphically as a parabola.

The vertex of the parabola occurs at #x=(-b)/(2a)#

i.e where #x=(-3)/-2 = 3/2 = 1.5#

Since #a<0# the vertex will be at the absolute maximum #f(x)#

#:.f_max = f(3/2) = -(3/2)^2+3(3/2)+10#

#= -9/4 + 9/2 +10 = 9/4+10 = 12.25#

Hence another 'important point' is: #f_max = (1.5, 12.25)#

We can see these points of the graph of #f(x)# below.

graph{-(x+2)(x-5) [-36.52, 36.52, -18.27, 18.27]}