Find the vertex and
intercepts of y
= 3x^{2} + x – 2
and graph;
remember to label the vertex and the axis of symmetry.

This is the same quadratic
as in the last example. I already found the vertex when I worked the
problem above. This time, I also need to find the intercepts
before I do my graph. To find the y-intercept,
I set x
equal to zero and solve:

y
= 3(0)^{2} + (0) – 2
= 0 + 0 – 2 = –2

Then the y-intercept
is the point (0,
–2). To find the
x-intercept,
I set y
equal to zero, and solve:

0 = 3x^{2}
+ x – 2
0
= (3x – 2)(x + 1)
3x
– 2 = 0 or x + 1 = 0
x
= ^{2}/_{3} or x = – 1

Then the x-intercepts
are at the points (–1,
0) and (
^{2}/_{3}, 0).

The axis of symmetry
is halfway between the two
x-intercepts
at (–1,
0) and at (
^{2}/_{3}
, 0); using
this, I can confirm the answer from the previous page:

(–1
+ 2/3) / 2 = (–1/3) / 2 = –1/6

The complete answer
is a listing of the vertex, the axis of symmetry, and all three
intercepts, along with a nice neat graph:

The vertex
is at ( ^{–1}/_{6}
, ^{–25}/_{12} ),
the axis of symmetry is the line x
= ^{–1}/_{6} ,
and
the intercepts are at (0,
–2), (–1,
0), and (
^{2}/_{3}, 0).

Find the intercepts,
the axis of symmetry, and vertex of y
= x^{2} – x – 12.

To find the y-intercept,
I set x
equal to 0
and solve:

y
= (0)^{2} – (0) – 12 = 0 – 0 – 12 = –12

To find the x-intercept,
I set y
equal to 0
and solve:

0 = x^{2}
– x – 12

0 = (x
– 4)(x + 3)

x
= 4 or x = –3

To find the vertex, I
look at the coefficients: a
= 1 and b
= –1. Plugging into
the formula, I get:

h =
^{–(–1)}/_{2(1)} = ^{1}/_{2} = 0.5

To find k,
I plug h
= ^{1}/_{2}
in for x
inside y
= x^{2} – x – 12,
and simplify:

The vertex is
at the point (0.5, –12.25),
the axis of symmetry is the line x
= 0.5, and
the intercepts are at the points (0,
–12), (–3,
0), and (4,
0).