Question

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

Answer 1

You'll have a parabola passing through:
`x=0, y=24`
`x=2, y=0`
`x=12, y=0`
and vertex at `x=7, y=-25`

Explanation:

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]}