Graph the following quadratic function.

f(x) = -x² – 2x + 6

Standard parabola graphing procedure applies here: find the vertex, the intercepts, and any additional points, then make the graph.

f(h) = -(-1)² – 2(-1) + 6 = -1 + 2 + 6 = 7

The vertex is (-1, 7).

y = -(0)² – 2(0) + 6 = 6

The y-intercept is (0, 6).

b² – 4ac = (-2)² – 4(-1)(6) = 4 + 24 = 28

There are two real roots, which means two x-intercepts. Better find 'em. Quadratic formula, where you at?

Beautiful. The x-intercepts are  and  And now we graph.

Graph the following quadratic function.

f(x) = x² + 3x + 5

Let's find the vertex, for starters.

The vertex is . Next up: intercepts.

f(0) = 0² + 3(0) + 5 = 5

The y-intercept is (0, 5).

b² – 4ac = 9 – 4(1)(5) = 9 – 20 = -11

The discriminant is negative, so there are no x-intercepts. You could also tell by seeing that the vertex is above the x-axis and a is positive, so the parabola points up.

We need a few extra points, so let's choose one from each side of the vertex.

f(-1) = (-1)² + 3(-1) + 5 = 1 – 3 + 5 = 3

f(-4) = (-4)² + 3(-4) + 5 = 16 – 12 + 5 = 9

x	f(x)
0	5
-1	3
-4	9
-3	5
-2	3
1	9

How'd we get those last three points? Like this: (-3, 5), (-2, 3), and (1, 9) are symmetric to (0, 5), (-1, 3), and (-4, 9).

Now graph away.

Graph the following parabola.

There are a lot of fractions in this problem, so be sure to take your time to avoid mistakes.

The vertex is .

The y-intercept is .

Because the vertex lies on the x-axis, we're guessing there will only be one x-intercept. We should check the discriminant just to be safe, though.

The discriminant is 0 so, as we thought, there's only one x-intercept.

We need a few more points to make the graph. We're going to use x = 1, -1, and 3 to try and simplify the fractions some.

x	y
	0
0
1
-1
3