How do you graph a non-factorable trinomial

Something like

x^2 - 2x - 1

or any other trinomial that you wouldn't be able to factor, how do I go about finding the x/y intercepts?

Well, the y-intercept is when x=0 so in this case it would be y = -1

The x-intercepts are when y=0 so if it does not factor easily then you can use the Quadratic Formula or Completing the Square.

I'll complete the square on this one:

x^2 - 2x - 1 = 0

x^2 - 2x = 1

x^2 - 2x + 1 = 1 + 1

(x-1)^2 = 2

x-1 = +/-sqrt(2)

x = 1 +/-sqrt(2)

:)

1. you graph by picking an x, and seeing what y becomes when you plug in the x value you choose

2. you graph intercepts by choosing the zero value for the other part. the y-intercept means that x is zero, and the same way the x-intercept means the y value is zero.

3. y = x^2 -2x -1.

A. (letting x = 0)
y = (0)^2 -2(0) -1.
y = 0 -0 -1.
y = -1.
(0,-1) exists on the line.

B. if you cannot solve for an exact answer you can pattern match, or even just fill in values you pick, and graph them, and see approximately the x intercept. (draw the line)
i.e. when does x^2 - 2x -1 go to zero.
(pick x as 1,etc and see what happens)
f(1) = (1)^2 - 2(1) - 1 = 1 -2 -1 = 1-3 = -2. (1,-2) is on the line.
f(2) = (2)^2 -2(2) -1 = 4 -4 -1 = 4 - 5= -1. (2,-1) is on the line.
notice how we are getting closer to the value of 0 for y?
f(3) = (3)^2 -2(3) - 1 = 9 - 6 -1 = 9 - 7 = 2. (3,2) is on the line.
and we passed the point at which the line crossed y=0 between x=2 and x = 3.

similarly we could try to find the other y intercept, on the left of x=0, y = -1.
f(-1) = (-1)^2 -2(-1) -1 = 1 +2 - 1 = 3-1 = 2. (-1,2) exist on the line.
and we passed the point at which the line crossed y=0 between x = 0 and x = 1.

you can further check if you want. (or use the quadratic formula but only if a binomial).