Chapter 1 Functions And Graphs Section 13 Exercises Answers
Are you struggling to find answers to the exercises in Chapter 1 Functions and Graphs Section 13? You are not alone. Many students find it difficult to solve the exercises and get the correct answers. In this article, we will provide you with step-by-step solutions to the exercises in Chapter 1 Functions and Graphs Section 13. We will use relaxed language to make it easier for you to understand.
Exercise 1
The first exercise in Section 13 asks you to find the slope of a line passing through two points. The two points are (3, 4) and (6, 8). To find the slope, we use the formula:
m = (y2 - y1) / (x2 - x1)
Substituting the values, we get:
m = (8 - 4) / (6 - 3) = 4 / 3
Therefore, the slope of the line passing through the two points is 4/3.
Exercise 2
The second exercise in Section 13 asks you to find the equation of a line passing through a point and parallel to another line. The point is (2, -3) and the line is y = 5x + 2. To find the equation of the line, we use the formula:
y - y1 = m(x - x1)
Since the line we are looking for is parallel to the given line, the slope of the line is the same as the slope of the given line. Therefore, the slope is 5. Substituting the values, we get:
y - (-3) = 5(x - 2)
Simplifying the equation, we get:
y = 5x - 13
Therefore, the equation of the line passing through the point (2, -3) and parallel to the line y = 5x + 2 is y = 5x - 13.
Exercise 3
The third exercise in Section 13 asks you to find the equation of a line passing through two points. The two points are (1, 3) and (4, 5). To find the equation of the line, we use the formula:
y - y1 = (y2 - y1) / (x2 - x1) (x - x1)
Substituting the values, we get:
y - 3 = (5 - 3) / (4 - 1) (x - 1)
Simplifying the equation, we get:
y = (2/3)x + 7/3
Therefore, the equation of the line passing through the two points (1, 3) and (4, 5) is y = (2/3)x + 7/3.
Exercise 4
The fourth exercise in Section 13 asks you to find the x- and y-intercepts of a line. The equation of the line is y = -2x + 6. To find the x-intercept, we set y to zero and solve for x:
0 = -2x + 6
x = 3
Therefore, the x-intercept of the line is 3. To find the y-intercept, we set x to zero and solve for y:
y = -2(0) + 6
y = 6
Therefore, the y-intercept of the line is 6.
Exercise 5
The fifth exercise in Section 13 asks you to find the domain and range of a function. The function is:
f(x) = x^2 - 4
To find the domain, we need to find all the values of x for which the function is defined. Since the function is a polynomial, it is defined for all real values of x. Therefore, the domain of the function is (-∞, ∞).
To find the range, we need to find all the values of y that the function can take. We can rewrite the function as:
f(x) = (x + 2)(x - 2)
Since the square of any real number is always positive or zero, the minimum value of f(x) is -4. Therefore, the range of the function is [-4, ∞).
Exercise 6
The sixth exercise in Section 13 asks you to find the inverse of a function. The function is:
f(x) = 2x + 1
To find the inverse of the function, we need to interchange the roles of x and y and solve for y:
x = 2y + 1
y = (x - 1) / 2
Therefore, the inverse of the function is:
f-1(x) = (x - 1) / 2
Exercise 7
The seventh exercise in Section 13 asks you to find the maximum and minimum values of a function. The function is:
f(x) = x^2 - 6x + 9
To find the maximum and minimum values, we complete the square:
f(x) = (x - 3)^2
Since the square of any real number is always positive or zero, the minimum value of f(x) is zero, which occurs when x = 3. Therefore, the minimum value of the function is 0.
Since the square of any real number is always positive or zero, the function has no maximum value. However, we can say that the function approaches infinity as x approaches positive or negative infinity.
Exercise 8
The eighth exercise in Section 13 asks you to find the x- and y-coordinates of the vertex of a parabola. The equation of the parabola is:
y = -2x^2 + 8x - 5
To find the x-coordinate of the vertex, we use the formula:
x = -b / 2a
Substituting the values, we get:
x = -8 / (-4) = 2
Therefore, the x-coordinate of the vertex is 2. To find the y-coordinate of the vertex, we substitute x = 2 in the equation of the parabola:
y = -2(2)^2 + 8(2) - 5
y = 3
Therefore, the y-coordinate of the vertex is 3.
Exercise 9
The ninth exercise in Section 13 asks you to find the x- and y-intercepts of a parabola. The equation of the parabola is:
y = 2x^2 - 4x + 1
To find the x-intercepts, we set y to zero and solve for x:
0 = 2x^2 - 4x + 1
Using the quadratic formula, we get:
x = (4 ± √12) / 4
x = (1 ± √3) / 2
Therefore, the x-intercepts of the parabola are (1 + √3)/2 and (1 - √3)/2.
To find the y-intercept, we set x to zero and solve for y:
y = 1
Therefore, the y-intercept of the parabola is 1.
Exercise 10
The tenth exercise in Section 13 asks you to find the domain and range of a parabola. The equation of the parabola is:
y = -3x^2 + 6x - 2
To find the domain, we need to find all the values of x for which the parabola is defined. Since the parabola is a polynomial, it is defined for all real values of x. Therefore, the domain of the parabola is (-∞, ∞).
To find the range, we need to find the maximum or minimum value of the parab
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