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Intermediate Algebra Review Notes Exam #4
Exercise 6(a) To place f(x) = x^{2} 6x10 in vertex form,
take onehalf 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 xintercept, 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 xintercepts 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 onehalf 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 yintercept is (0, 12). To find the xintercept, 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 xintercepts 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
yvalue 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
2x5 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 xvalue 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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