How do you graph #y=x^2-14x+24#?

This is a Quadratic and its graph will be a Parabola (kind of "U" shaped curve). To plot it you may consider some interesting features of your equation:

1] the coefficient of #x^2# is #1>0# so yours will be a "upward" or "happy" parabola (in the shape of a smiling "U");

2] set #x=0# into your equation to find the #y#-intercept:
#y=0+0+24#
so your parabola will cross the y axis at #y=24#;

3] set #y=0# into your equation to find (if they exists) the #x#-intercepts(s):
#x^2-14x+24=0# solving using the Quadratic Formula you get:
#x_(1,2)=(14+-sqrt(196-96))/2==(14+-10)/2=#
you get two values:
#x_1=24/2=12#
#x_2=4/2=2#
so your parabola crosses the x axis at #x=2 and x=12#;

4] the vertex; this is a very important point because it characterize the entire graph setting the central point you need to have when plotting a parabola.
Given the equation in general form #y=ax^2+bx+c#; in your case you have:
#a=1#
#b=-14#
#c=24#
The coordinates of the vertex are then given as:
#x_v=-b/(2a)=-(-14)/2=7#
#y_v=-Delta/(4a)=-(b^2-4ac)/(4a)=-100/4=-25#

Finally your graph will look like:
graph{x^2-14x+24 [-83.3, 83.3, -41.64, 41.64]}