# Intermediate Algebra Problems With Answers - Sample 4

A set of intermediate algebra problems on functions, domain, range, collinear points, zeros of functions, x and y intercepts, ... with answers, are presented. The solutions are at the bottom of the page.

1. Find the constant b so that the three points A(2 , 3), B(4 , 7) and C(8 , b) are collinear (i.e. on the same line).

2. Which of the following equations represent y as a function of x?
a) x 2 + y = 5
b) x 2 - 7x + y 2 = 0
c) x 2 - y 3 - 9 = 0
d) |x| - |y| = 0

3. Which set of ordered pairs represents a function?
a) A = { (a , 3) , (b , 5) , (c , 9) , (d , 9) }
b) B = { (a , -3) , (b , 6) , (c , 1) , (b , 9) }
c) C = { (a , 3) , (b , 3) , (c , 3) , (d , 3) }
d) D = { (a , 5) , (a , -9) , (a , 0) , (a , 12) }

4. Find the domain of each function.
a) f(x) = 1 / (x - 2)
b) g(x) = √(x + 5)
c) h(x) = 3 / √(x - 4)
d) j(x) = 1 / [ (x + 1)(x -7) ]

5. Find the zeros of each function.
a) f(x) = 3 - 5/x
b) g(x) = √(x - 1)
c) h(x) = | x - 9 | - 4
d) i(x) = | x - 9 | + 7
d) j(x) = x 2 - 16
d) k(x) = x 2 + 3

6. Find the range of each function.
a) f(x) = x 2
b) g(x) = | x |
c) h(x) = x 2 + 6
d) j(x) = | x | + 2

7. Let f(x) = 2x 2 - 4x + 4 and h(x) = 2x - 4. Find
a) f(2) =
b) h(5) =
c) f(3) + h(3) =

8. Find the x and y intercepts of the graphs of the following equations.
a) 2x + 4y = 5
b) x 2 + (y - 3) 2 = 9
c) |x - 3| + |5 - y| = 6

9. Find the linear function f(x) = A x + B such that f(2) = 1 and f(4) = -3.

10. Which of these functions is even?
a) f(x) = -x 2 + 7
b) g(x) = | x - 6 |
c) h(x) = x 3 + 9
d) j(x) = | x | + 1

Detailed Solutions to the Above Questions

1. The slope of the line through A and B must be equal to the slope of the line through B and C.
(7 - 3)/(4- 2) = (b - 7)/(8 - 4)
solve the above for b to find b = 15.

2. a) solve for to find: y = 5 - x 2 , for each value of x there is only one corresponding value of y. y is a function of y.
b) Solve for y to find: y = ~+mn~√(7x - x 2) , there are values of x that give two values of y and therefore y is not a function of x.
c) solve for y to find: y 3 = ( x 2 - 9 ) 1/3 , for each value of x there is one value of y and therefore y is a function of x.
d) solve for y to find: y = ~+mn~x , there are values of x that give two values of y and therefore y is not a function of x.

3. a) Relation A is a function. For each x (first value) value corresponds only one value of y (second value).
b) Relation B is not a function. The two pairs (b , 6) and (b , 9) means that for x = b there are two different values of y: 6 and 9.
c) Relation C is a function. For each x value corresponds only one value of y.
d) Relation D is not a function. For x = a, y has different values.

4. a) f(x) = 1 / (x - 2) , division by zero is not allowed hence x - 2 must be different from 0.
domain (-∞ , 2) U (2 , ∞)
b) g(x) = √(x + 5) , for g(x) to be real x + 5 must be greater then or equal to 0. Hence x + 5 ≥ 0
domain: [-5 , + ∞)
c) h(x) = 3 / √(x - 4) , for h(x) to be real x - 4 must be positive.
domain: (4 , + ∞)
d) j(x) = 1 / [ (x + 1)(x - 7) ] , division by 0 is not allowed hence x cannot be equal to -1 or 7.
domain: (-∞ , - 1) ∪ (-1 , 7) ∪ (7 , + ∞)

5. The zeros of a given function f are the solution(s) to the equation f(x) = 0.
a) 3 - 5/x = 0 , zero at x = 5/3
b) √(x - 1) = 0 , zero at x = 1
c) | x - 9 | - 4 = 0 , zeros at x = 13 and x = 5
d) | x - 9 | + 7 = 0 , no zeros
e) x 2 - 16 = 0 , zeros at x = 4 and x = -4
f) x 2 + 3 = 0 , no zeros.

6. The range of a given function f is the set of values of f(x) for x in the domian of f.
a) f(x) = x 2. The square of a real number x is either positive or zero for x = 0, hence x 2 ≥ 0. The domain of f is given by the interval [0 , + ∞).
b) g(x) = | x |. The absolute value of a real number x is either positive or zero for x = 0, hence | x | ≥ 0. The domain of g is given by the interval [0 , + ∞).
c) h(x) = x 2 + 6. We have seen above that x 2 ≥ 0. Add 6 to both sides of the inequality to obtain x 2 + 6 ≥ 6. The left side of the inequality is h(x). Hence h(x) ≥ 6 and the domain of h is given by the interval [6 , + ∞).
c) j(x) = | x | + 2. We have seen above that | x | ≥ 0. Add 2 to both sides of the inequality to obtain | x | + 2 ≥ 2. The left side of the inequality is j(x). Hence j(x) ≥ 2 and the domain of j is given by the interval [2 , + ∞).
7. f(x) = 2x 2 - 4x + 4 and h(x) = 2x - 4.
a) f(2) = 2(2) 2 - 4(2) + 4 = 4
a) h(5) = 2(5) - 4 = 6
a) f(3) + h(3) = 2(3) 2 - 4(3) + 4 + 2(3) - 4 = 12

8. a) 2x + 4y = 5. To find the x intercept set y = 0 in the given equation and solve for x.
2x + 4(0) = 5 , x = 5/2 , x intercept (5/2 , 0)
To find the y intercept set x = 0 in the given equation and solve for y.
2(0) + 4y = 5 , y = 5/4 , y intercept (0 , 5/4)
b) x 2 + (y - 3) 2 = 9
set y = 0 and solve for x: x 2 + (0 - 3) 2 = 9
x = 0 , x intercept (0 , 0)
set x = 0 and solve for y: 0 2 + (y - 3) 2 = 9
y = 0 , y = 6 ; 2 y intercpets: (0 , 0) and (0 , 6)
c) |x - 3| + |5 - y| = 6
y = 0, |x - 3| + |5 - 0| = 6, solve for x: x = 4 , x = 2.
x intercepts: (4 , 0) and (2 , 0)
x = 0, |0 - 3| + |5 - y| = 6, solve for y: y = 8 , y = 2
y intercepts: (0 , 8) and (0 , 2)
9. f(x) = A x + B, f(2) = 1 and f(4) = -3.
f(2) = 1 gives the equation: 1 = 2A + B
f(4) = -3 gives the equation: -3 = 4A + B
Solve the system of equations: 1 = 2A + B and -3 = 4A + B to find A = -2 and B = 5
Hence f(x) = -2x + 5
10. If a given function f is such that f(-x) = f(x), then f is an even function.
a) f(x) = -x 2 + 7 , f(-x) = - (- x) 2 + 7 = -x 2 + 7. Function f is an even function.
b) g(x) = | x - 6 |, g(-x) = | - x - 6 | = |-(x + 6) | = |x + 6| . Function g is not even.
c) h(x) = x 3 + 9, h(-x) = (-x) 3 + 9 = -x 3 + 9. Function h is not even.
d) j(x) = | x | + 1, j(-x) = | -x | + 1 = | x | + 1. Function j is even.

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