# 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.

 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 Detailed Solutions 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 (-∞ , 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 , + ∞) 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 , + ∞). 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 , + ∞). 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. Algebra Questions and problems More ACT, SAT and Compass practice