EJU - Examination for Japanese University Mathematics: Functions and Graphs Questions and Answers

Question 1

Find the coordinates of the vertex of the parabola defined by the quadratic function y = 2x² - 8x + 5.

Accepted Answer

(2, -3)

Answer

The x-coordinate of the vertex of a parabola y = ax² + bx + c is given by the formula x = -b / 2a. For the function y = 2x² - 8x + 5, a = 2 and b = -8. Thus, x = -(-8) / (2 * 2) = 8 / 4 = 2. To find the y-coordinate, substitute x = 2 back into the equation: y = 2(2)² - 8(2) + 5 = 2(4) - 16 + 5 = 8 - 16 + 5 = -3. Therefore, the vertex is at (2, -3).

Question 2

The graph of the function y = f(x) is translated 3 units to the left and 4 units down. Which of the following equations represents the new graph?

Accepted Answer

y = f(x + 3) - 4

Answer

A horizontal translation of h units is represented by f(x - h), and a vertical translation of k units is represented by f(x) + k. A translation 3 units to the left corresponds to h = -3, so the term becomes f(x - (-3)) = f(x + 3). A translation 4 units down corresponds to k = -4. Combining these gives the new function y = f(x + 3) - 4.

Question 3

What is the period of the function y = 3sin(2x - π/2)?

Accepted Answer

π

Answer

The period of a trigonometric function of the form y = A sin(Bx + C) + D is given by the formula Period = 2π / |B|. In the function y = 3sin(2x - π/2), the value of B is 2. Therefore, the period is 2π / 2 = π.

Question 4

Which of the following statements is true about the relationship between the graph of a function y = f(x) and its inverse function y = f⁻¹(x)?

Accepted Answer

They are reflections of each other across the line y = x.

Answer

The graph of a function and its inverse are always symmetric with respect to the line y = x. This is because if a point (a, b) is on the graph of f(x), then the point (b, a) must be on the graph of f⁻¹(x). The line y = x is the perpendicular bisector of the segment connecting these two points.

Question 5

A logarithmic function is given by y = log₂(x - 3). What is the domain of this function?

Accepted Answer

x > 3

Answer

The argument of a logarithmic function must be strictly positive. For the function y = log₂(x - 3), this means that the expression (x - 3) must be greater than 0. Solving the inequality x - 3 > 0 gives x > 3. Therefore, the domain of the function is all real numbers greater than 3.

EJU - Examination for Japanese University Practice Test

EJU - Examination for Japanese University Practice Test

EJU - Examination for Japanese University Mathematics: Functions and Graphs Questions and Answers