Intermediate Algebra Problems With Answers  Sample 4
A set of intermediate algebra problems, with answers, are presented. The solutions are at the bottom of the page.

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).

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

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) }

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) ]

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

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

Let f(x) = 2x^{ 2}  4x + 4 and h(x) = 2x  4. Find
a) f(2) =
b) h(5) =
c) f(3) + h(3) =

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

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

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
Answers to the Above Questions

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.

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.

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.

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

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.

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 , + infinity).
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 , + infinity).
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 , + infinity).
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 , + infinity).

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

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)

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

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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