**Example 11.** Complete the square to place f(x) =
2x^{2} + 4x − 4 in vertex form

and sketch its graph.

In the last two examples, we gained some measure of
success when the coefficient

of x^{2} was a 1. We were just getting comfortable with that situation and we’ d
like

to continue to be comfortable, so let’s start by factoring a 2 from each term on
the

right -hand side of the equation .

f(x) = 2[x^{2} + 2x − 2]

If we ignore the factor of 2 out front, the coefficient of x^{2} in the trinomial
expression

inside the parentheses is a 1. Ah, familiar ground! We will proceed as we did
before,

but we will carry the factor of 2 outside the parentheses in each step . Start by
taking

half of the coefficient of x and squaring the result; i.e., [(1/2)(2)]^{2} = 1.

Add and subtract this amount inside the parentheses so as
to not change the equation.

f(x) = 2[x^{2} + 2x + 1 − 1 − 2]

Group the first three terms inside the parentheses and combine constants .

f(x) = 2[(x^{2} + 2x + 1) − 3]

The grouped terms inside the parentheses form a perfect square trinomial that is
easily

factored.

f(x) = 2[(x + 1)^{2} − 3]

Finally, redistribute the 2.

f(x) = 2(x + 1)^{2} − 6

This is a parabola that opens upward. In addition, it is
stretched by a factor of

2, so it will be somewhat narrower than our previous examples . The parabola is
also

shifted 1 unit to the left, then 6 units downward, placing the vertex at
(−1,−6), as

shown in Figure 3(a). The table in Figure 3(b) calculates two points to the
right of

the axis of symmetry, and mirror points on the left of the axis of symmetry make
for

an accurate plot of the parabola.

Figure 3. Plotting the graph of the

quadratic function f(x) = 2x^{2} + 4x − 4.

Let’s look at an example where the coefficient of x^{2} is
negative .

**Example 12.** Complete the square to place f(x) =
−x^{2} + 6x − 2 in vertex form

and sketch its graph.

In the last example, we factored out the coefficient of
x^{2}. This left us with a

trinomial having leading coefficient 1, which enabled us to proceed much as we
did

before: halve the middle coefficient and square, add and subtract this amount,
factor

the resulting perfect square trinomial. Since we were successful with this
technique in

the last example, let’s begin again by factoring out the leading coefficient, in
this case

a −1.

f(x) = −1[x^{2} − 6x + 2]

If we ignore the factor of −1 out front, the coefficient of x^{2} in the trinomial
expression

inside the parentheses is a 1. Again, familiar ground! We will proceed as we did
before,

but we will carry the factor of −1 outside the parentheses in each step. Start
by taking

half of the coefficient of x and squaring the result; i.e., [(1/2)(−6)]^{2} = 9.

Add and subtract this amount inside the parentheses so as
to not change the equation.

f(x) = −1[x^{2} − 6x + 9 − 9 + 2]

Group the first three terms inside the parentheses and combine constants.

f(x) = −1[(x^{2} − 6x + 9) − 7]

The grouped terms inside the parentheses form a perfect square trinomial that is
easily

factored.

f(x) = −1[(x − 3)^{2} − 7]

Finally, redistribute the −1.

f(x) = −(x − 3)^{2} + 7

This is a parabola that opens downward. The parabola is
also shifted 3 units to

the right, then 7 units upward, placing the vertex at (3, 7), as shown in Figure
4(a).

The table in Figure 4(b) calculates two points to the right of the axis of
symmetry,

and mirror points on the left of the axis of symmetry make for an accurate plot
of the

parabola.

Figure 4. Plotting the graph of the

quadratic function f(x) = −(x − 3)^{2} + 7.

Let’s try one more example .

**Example 13.** Complete the square to place f(x) =
3x^{2} + 4x − 8 in vertex form

and sketch its graph.

Let’s begin again by factoring out the leading
coefficient, in this case a 3.

Fractions add a degree of difficulty, but, if you fol low
the same routine as in the previous

examples, you should be able to get the needed result. Take half of the
coefficient of x

and square the result; i.e., [(1/2)(4/3)]^{2} = [2/3]^{2} = 4/9.

Add and subtract this amount inside the parentheses so as
to not change the equation.

Group the first three terms inside the parentheses. You’ll
need a common denominator

to combine constants.

The grouped terms inside the parentheses form a perfect
square trinomial that is easily

factored.

Finally, redistribute the 3.

This is a parabola that opens upward. It is also stretched
by a factor of 3, so it

will be narrower than all of our previous examples. The parabola is also shifted
2/3

units to the left, then 28/3 units downward, placing the vertex at (−2/3,−28/3),
as

shown in Figure 5(a). The table in Figure 5(b) calculates two points to the
right of

the axis of symmetry, and mirror points on the left of the axis of symmetry make
for

an accurate plot of the parabola.

Figure 5. Plotting the graph of the quadratic

function f(x) = 3(x + 2/3)^{2} − 28/3.