# Intermediate Algebra Review Notes Exam #4

**Exercise 6(a) **To place f(x) = x^{2} -6x-10 in vertex form,
take one-half of the middle coefficient

and square, i.e., [(1/2)(-6)] = 9. Add and subtract this amount , factor and
simplify .

f(x) = x^{2} - 6x + 9 - 9 - 10

f(x) = (x - 3)^{2} - 19

The parabola opens upward, the vertex is at (3,-19), and the equation of the
axis of symmetry is

x = 3.

Because f(0) = -10, the y- intercept is (0,-10). To find the x-intercept, set y =
0.

0 = x^{2} - 6x - 10

Note that ac = (1)(-10) = -10. There is no integer pair that has product -10 and
sums to -6.

Hence, we will need to use the quadratic formula .

Thus, the x-intercepts areand
. Using a calculator , these approximately

equal -1.35 and 7.35.

**Exercise 6(b)** To place f(x) = -x^{2} - 5x + 12 in vertex
form, first factor out a -1.

f(x) = - [
x^{2} + 5x - 12 ]

Take one-half of the middle coefficient and square , i.e., [(1/2)(5)] = 25/4. Add
and subtract this

amount, factor and simplify .

The parabola opens downward, the vertex is at (-5/2,
73/4), and the equation of the axis of

symmetry is x = -5/2.

Because f(0) = 12, the y-intercept is (0, 12). To find the x-intercept, set y = 0,
then multiply

both sides of the equation by -1.

0 = -x^{2} - 5x + 12

0 = x^{2} + 5x - 12

Note that ac = (1)(-12) = -12. There is no integer pair that has product -12 and
sums to 5.

Hence, we will need to use the quadratic formula .

Thus, the x-intercepts are
and . Using a calculator, these approxi-

mately equal -6.7720 and 1.7720.

**Exercise 7.** Let x and y represent numbers. Their difference
is 12, so

x - y = 12. (1)

The sum of their squares is

S = x^{2} + y^{2}. (2)

Solve equation (1) for x.

x = y + 12 (3)

Substitute equation (3) into equation (2), expand and simplify.

S = (y + 12)^{2} + y^{2}

S = y^{2} + 24y + 144 + y^{2}

S = 2y^{2} + 24y + 144

This is a parabola that opens upward, so the minimum S- value will occur at the
vertex. The

y-value of the vertex is given by

To find the second number, substitute y = -6 in equation
(3).

x = -6 + 12

x = 6

Thus, the numbers are x = 6 and y = -6.
Exercise 7

**Exercise 8**. One number is 5 less than twice a second number. So, the numbers are
2x-5 and x.

The product of the two numbers is given by

P = (2x - 5)x

P = 2x^{2} - 5x.

This is a parabola that opens upward, so P will have a minimum at the vertex of
the parabola.

The x-value of the vertex is given by

The first number is

Hence, the two numbers are 5/4 and -5/2.

**Solutions to Multiple Choice Questions
**

**Solution to Question 1:**-3

**Solution to Question 2:**1

**-x + 1**

Solution to Question 3:

Solution to Question 3:

**Solution to Question 4:**2x

**Solution to Question 5:**(-7, 7)

**Solution to Question 6:**(-∞,-7] ∪ ∪7,∞)

**Solution to Question 7:**No solutions

**Solution to Question 8:**x = 2

**Solution to Question 9:**(1/4,-9/8)

**Solution to Question 10:**

**Solution to Question 11:**None of these. The correct solution is k = -25/24.

**7**

Solution to Question 12:

Solution to Question 12:

**Solution to Question 13:**1.5

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