Example Question - line graph
Here are examples of questions we've helped users solve.
Finding Equation of a Line and Inequalities
To find the equation of the line, you need to find its slope and y-intercept. From the graph, you can see the line crosses the y-axis at the point (0, 1), which means the y-intercept is 1. To find the slope, you can pick two points on the line. The line passes through the points (0, 1) and (-2, -1) which can be clearly seen on the graph. The slope (m) is calculated as the rise over the run: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 1}{-2 - 0} = \frac{-2}{-2} = 1. \] So, the slope is 1. With a slope of 1 and a y-intercept of 1, the equation of the line in slope-intercept form (y = mx + b) is: \[ y = 1x + 1 \] or simply: \[ y = x + 1. \] For inequalities: a) Since the line is solid, this means that the inequality includes equality; that is, points on the line satisfy the inequality. Since the area below the line is shaded, the inequality represents all points where y is less than or equal to x + 1. Therefore, the inequality for part a) is: \[ y \leq x + 1. \] b) and c) are not provided explicitly in the question, but typically they would be related to different shading regions if provided. If the shaded region was above the line for part b), for example, the inequality would be y ≥ x + 1. If the shaded region for part c) omitted points on the line, the inequality would use strict inequality signs (< or >) rather than ≤ or ≥.
Calculating Slope and Y-Intercept from a Line Graph
To find the slope of the line represented in the graph, you need to pick two points on the line that you can clearly identify the coordinates for. Then, use the formula for slope, which is: \[ slope (m) = \frac{change \, in \, y}{change \, in \, x} = \frac{y_2 - y_1}{x_2 - x_1} \] Looking at the graph, let's pick the two points where the line intersects the grid lines. One point could be at \( (1, 1) \) and another point could be at \( (3, 7) \). Now using these points: \[ x_1 = 1, y_1 = 1 \] \[ x_2 = 3, y_2 = 7 \] Let's calculate the slope: \[ slope (m) = \frac{7 - 1}{3 - 1} = \frac{6}{2} = 3 \] Next, to find the y-intercept (the point where the line crosses the y-axis), look at where the line intersects the y-axis. From the graph, it appears that the line crosses the y-axis at \( (0, -1) \). Therefore, the y-intercept (b) is -1. So the slope is 3, and the y-intercept is -1.
Modeling Temperature Changes After Daylight Saving Time Shift
The image shows a line graph with the x-axis labeled "Hours" from 0 to 24 and the y-axis labeled "Temperature" from 50 to 90 degrees in increments of 10. The line graph itself indicates changes in temperature over the course of a day. The accompanying text describes that during a month in a certain town, the temperature increases and decreases over the course of a day as shown in the graph. A particular point is mentioned where the temperature readjusts due to time change for daylight saving time. If the clock is adjusted forward by one hour at 1 a.m., then what was originally 1 a.m. instantaneously becomes 2 a.m. The question stipulates that if we have a function T for temperature with respect to hours, we are seeking a function D that models the same data, considering the time shift for daylight saving time. To solve for function D in terms of function T after the time shift, if the function T(h) represents the temperature at hour h, then for the time after the time shift, we need to subtract one hour from the h values because the clock jumps forward one hour (i.e., what was 1 a.m. becomes 2 a.m.). Therefore, the point on the graph that was originally at T(h) is now at T(h-1) when h is after 1 a.m. Thus, if we denote D(h) as the function for temperature after the time shift, we can define D(h) in terms of T as: D(h) = T(h-1) for h > 1. Remember that this adjustment is only applicable for hours after the time shift occurs. For the first hour (from midnight to 1 a.m.), there is no shift, so D(h) would equal T(h) within that interval. However, since the question does not ask for this specific interval and from the context of the question it is clear that we're only considering the hours after the daylight saving time change, the general solution for hours after 1 a.m. would be as stated above.
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