1-2) See the picture above.
3) The slope of the line that passes through the points (x1, y1) and (x2, y2) is computed as follows:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
The initial climb passes through the points (0,0) and (20, 50), then its slope is:
[tex]m=\frac{50-0}{20-0}=\frac{5}{2}=2.5[/tex]
4) The equation of a line in slope-intercept form is:
y = mx + b
where m is the slope and b is the y-intercept
The line of the initial climb intersects the y-axis at the point (0,0), then b = 0.
Substituting with m = 5/2 and b = 0, the equation of the line of the initial climb is:
[tex]\begin{gathered} y=\frac{5}{2}x+0 \\ y=\frac{5}{2}x \end{gathered}[/tex]
5) The domain is the set of all possible x-values.
The range is the set of all possible y-values.
From the graph:
Domain: [0, 230]
Range: [0, 50
6) The rate of change of a function f(x) between the points x = a, and x = b, is computed as follows:
[tex]\text{ rate of change =}\frac{f(b)-f(a)}{b-a}[/tex]
Considering the points (50, 0) and (70, 35), which means
a = 50
b = 70
f(a) = 0
f(b) = 35
the rate of change of the first hill is:
[tex]\text{ rate of change =}\frac{35-0}{70-50}=\frac{7}{4}[/tex]
Considering the points (170, 0) and (200, 20), which means
a = 170
b = 200
f(a) = 0
f(b) = 20
the rate of change of the second hill is:
[tex]\text{ rate of change =}\frac{20-0}{200-170}=\frac{2}{3}[/tex]
The steeper hill is the first hill because its rate of change is greater.
7) In a function, for every x-value in the domain there is associated only 1 value of y in the range. We can see in the graph that x = 130 is associated with two y-values: y = 25 and y = 3.8. In consequence, the roller coaster is not a function.
