University:
Carleton College
Course:
MATH 120 | Calculus 2
Academic year:

2024
Views:

323

Pages:

2
Author:

Nevaeh Bauer

Question 1 Determine if each relationship is an additive relationship. Explain. Input(x) | 4 | 5 | 7 | 9 | 10 | 13 | Output(y) | 5 | 7 | 9 | 10 | 13 | Answer The relationship is additive if you can add the same number to each x-value to get the y- values. For the first pair of x = 4 and y = 5, you need to add 1 since 4 + 1 = 5. For the second pair of x = 5 and y = 7, you need to add 2 since 5 + 2 = 7. Since you need to add different numbers for these pairs, then the relationship is not additive. Question 2 The graph is a translation of the graph of f(x) = 2x². Write the function for the graph in vertex form. [Image of a graph with a parabola opening upwards. The vertex of the parabola is at (1, -3)] Answer Vertex form is provided by: f(x) = a(x-h)² + k where the (h,k) is the vertex. The vertex being at (1, - 3), the function for the graph is: g(x) = 2(x-1)² + (-3) or g(x) = 2(x-1)² - 3 Question 3 Use the piecewise-defined function to fill in the bla f(x) = {4, -4 < x < -2; 2x - 4, -1 < x < 2; 3x, 2 ≤ x ≤ 5} The domain ________ is used when graphing the function f(x) = 2x - 4. The equation ________ is used to find f(4) = 12. Answer The x values constitute the domain, so determining the domain for graphing f(x) = 2x - 4, you need to determine what values this piece is defined for. Derived from the partially defined function f(x) = 2x - 4 is defined for -1 < x < 2 If you are finding f(4), then x = 4. The domain that contains x = 4 is derived from the piecemeal defined function. is 2 ≤ x ≤ 5. This domain is for the piece of f(x) = 3x. Thus, the equation f(x) = 3x is used to find f(4) = 12. Question 4 Solve y=f(x) for x. Then find the input(s) when the output is -3. f(x)=3x+5 Answer Let y = f(x): y = 3x + 5 y - 5 = 3x + 5 - 5 y - 5 = 3x (y - 5)/3 = x So, y = -3, x = (-3 - 5)/3 x = -8/3 Question 5 Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros. Write the polynomial in standard form. -2, 1, 3 Answer If a is a zero of a polynomial, then x-a is one of its factors. Given the zeros -2, 1, and 3, we write: f(x)=a(x+2)(x-1)(x-3) Expand: f(x) = a(x² + x - 2)(x - 3) f(x) = a[x(x²+x-2) - 3(x²+x-2)] f(x) = a(x³+x²-2x -3x² -3x+6) f(x) = a(x³ -2x² - 5x + 6) Since the leading coefficient is 1, we let a=1: f(x) = x³ - 2x² - 5x + 6