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# Intermediate Algebra Review Notes Exam #4

Exercise 6(a) To place f(x) = x2 -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) = x2 - 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 = x2 - 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 are and . Using a calculator , these approximately
equal -1.35 and 7.35. Exercise 6(b) To place f(x) = -x2 - 5x + 12 in vertex form, first factor out a -1.

f(x) = - [ x2 + 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 = -x2 - 5x + 12
0 = x2 + 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
= x2 + y2. (2)

Solve equation (1) for x.

x = y + 12 (3)

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

S = (y + 12)2 + y2
S = y2 + 24y + 144 + y2
S = 2y2 + 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 = 2x2 - 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

Solution to Question 3:
-x + 1

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.

Solution to Question 12:
7

Solution to Question 13: 1.5

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