Question

write an equation in slope - intercept form for each graph shown.

1. graph with points (1, 0), (0, - 2)
2. graph with points (0, 3), (3, 0)
3. graph with points (4, - 2), (0, - 6)
4. graph (description not fully clear)

5a. enter the initial value:
b. enter the rate of change:
c. enter the equation of the line: graph with x - axis 0 - 20 and y - axis 0 - 20
6a. enter the initial value:
b. enter the rate of change:
c. enter the equation of the line: drew’s scuba tank graph with x - axis (minutes) and y - axis (tank level)

1. graph with points (2, 1), (0, - 3)
2. graph with points (0, 3), (2, - 3)
3. graph with points (0, - 1), (2, - 3)
4. tank temperature

graph with x - axis time (h) 0 - 5 and y - axis temperature (°f) 72 - 82
a. enter the initial value:
b. enter the rate:
c. enter the equation of the line:

1. this graph represents a linear function.

graph with x - axis and y - axis
a. enter the initial value:
b. enter the rate:
c. enter the equation of the line:

Answer:

Let's solve problem 1 (writing the equation in slope - intercept form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept) as an example.

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0, - 2)\). In the slope - intercept form \(y=mx + b\), when \(x = 0\), \(y=b\). So, \(b=-2\).

Step 2: Calculate the slope (\(m\))

We also know that the graph passes through the point \((1,0)\). The formula for slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Let \((x_1,y_1)=(0, - 2)\) and \((x_2,y_2)=(1,0)\). Then \(m=\frac{0-(-2)}{1 - 0}=\frac{0 + 2}{1}=2\).

Step 3: Write the equation

Using the slope - intercept form \(y=mx + b\) with \(m = 2\) and \(b=-2\), we get \(y = 2x-2\).

Let's solve problem 2:

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0,3)\), so \(b = 3\).

Step 2: Calculate the slope (\(m\))

The graph also passes through the point \((3,0)\). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) with \((x_1,y_1)=(0,3)\) and \((x_2,y_2)=(3,0)\), we have \(m=\frac{0 - 3}{3-0}=\frac{-3}{3}=-1\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m=-1\) and \(b = 3\), we get \(y=-x + 3\).

Let's solve problem 3:

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0,-6)\), so \(b=-6\).

Step 2: Calculate the slope (\(m\))

The graph also passes through the point \((4,-2)\). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) with \((x_1,y_1)=(0,-6)\) and \((x_2,y_2)=(4,-2)\), we get \(m=\frac{-2-(-6)}{4 - 0}=\frac{-2 + 6}{4}=\frac{4}{4}=1\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m = 1\) and \(b=-6\), we get \(y=x-6\).

Let's solve problem 7:

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0,-3)\), so \(b=-3\).

Step 2: Calculate the slope (\(m\))

The graph also passes through the point \((2,1)\). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) with \((x_1,y_1)=(0,-3)\) and \((x_2,y_2)=(2,1)\), we have \(m=\frac{1-(-3)}{2-0}=\frac{1 + 3}{2}=\frac{4}{2}=2\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m = 2\) and \(b=-3\), we get \(y = 2x-3\).

Let's solve problem 8:

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0,3)\), so \(b = 3\).

Step 2: Calculate the slope (\(m\))

The graph also passes through the point \((2,-1)\). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) with \((x_1,y_1)=(0,3)\) and \((x_2,y_2)=(2,-1)\), we get \(m=\frac{-1 - 3}{2-0}=\frac{-4}{2}=-2\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m=-2\) and \(b = 3\), we get \(y=-2x + 3\).

Let's solve problem 9:

Step 1: Identify the y - intercept (\(b\))

The graph passes through the point \((0,-1)\), so \(b=-1\).

Step 2: Calculate the slope (\(m\))

The graph also passes through the point \((2,-3)\). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\) with \((x_1,y_1)=(0,-1)\) and \((x_2,y_2)=(2,-3)\), we have \(m=\frac{-3-(-1)}{2-0}=\frac{-3 + 1}{2}=\frac{-2}{2}=-1\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m=-1\) and \(b=-1\), we get \(y=-x-1\).

Let's solve problem 10 (Tank Temperature):

Step 1: Identify the initial value (y - intercept, \(b\))

The graph passes through the point \((0,82)\), so the initial value (A) is \(82\).

Step 2: Calculate the rate (slope, \(m\))

We can take two points, say \((0,82)\) and \((1,80)\) (from the graph). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we have \(m=\frac{80 - 82}{1-0}=\frac{-2}{1}=-2\). So the rate (B) is \(-2\) (since the temperature is decreasing, the rate is negative).

Step 3: Write the equation

Using \(y=mx + b\) with \(m=-2\) and \(b = 82\), we get \(y=-2x + 82\) (C).

Let's solve problem 11 (the V - shaped graph, a piecewise linear function, but we can consider the right - hand side for the linear part starting from \(x = 0\)):

Step 1: Identify the initial value (y - intercept, \(b\))

The graph passes through the point \((0,-6)\), so the initial value (A) is \(-6\).

Step 2: Calculate the rate (slope, \(m\))

We can take two points, say \((0,-6)\) and \((2,0)\) (from the graph). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we have \(m=\frac{0-(-6)}{2-0}=\frac{6}{2}=3\). So the rate (B) is \(3\) (for \(x\geq0\)).

Step 3: Write the equation

Using \(y=mx + b\) with \(m = 3\) and \(b=-6\) (for \(x\geq0\)), we get \(y = 3x-6\) (C, for \(x\geq0\); the left - hand side will have a slope of \(-3\) to make the V - shape, but for the part starting at \(x = 0\), this is the equation).

For problem 5 (the first graph in the middle):

Step 1: Identify the initial value (y - intercept, \(b\))

The graph starts at \((0,2)\) (from the y - axis), so the initial value (5A) is \(2\).

Step 2: Calculate the rate (slope, \(m\))

We can take two points, say \((0,2)\) and \((2,4)\) (estimating from the graph). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we have \(m=\frac{4 - 2}{2-0}=\frac{2}{2}=1\). So the rate of change (5B) is \(1\).

Step 3: Write the equation

Using \(y=mx + b\) with \(m = 1\) and \(b = 2\), we get \(y=x + 2\) (5C).

For problem 6 (Drew's Scuba Tank):

Step 1: Identify the initial value (y - intercept, \(b\))

From the graph, when \(x = 0\) (time = 0), the amount of air (or whatever is measured on the y - axis) is, say, let's assume the initial point is \((0,10)\) (from the graph). So the initial value (6A) is \(10\) (you can estimate more accurately from the graph, but let's assume).

Step 2: Calculate the rate (slope, \(m\))

We can take two points, say \((0,10)\) and \((6,0)\) (from the graph). Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), we have \(m=\frac{0 - 10}{6-0}=\frac{-10}{6}=-\frac{5}{3}\approx - 1.67\). So the rate of change (6B) is \(-\frac{5}{3}\) (since the amount is decreasing).

Step 3: Write the equation

Using \(y=mx + b\) with \(m=-\frac{5}{3}\) and \(b = 10\), we get \(y=-\frac{5}{3}x + 10\) (6C).

Final Answers (selected problems):

1. \(y = 2x-2\)
2. \(y=-x + 3\)
3. \(y=x-6\)
4. \(y = 2x-3\)
5. \(y=-2x + 3\)
6. \(y=-x-1\)
7. A: \(82\); B: \(-2\); C: \(y=-2x + 82\)
8. A: \(-6\); B: \(3\); C: \(y = 3x-6\) (for \(x\geq0\))
9. A: \(2\); B: \(1\); C: \(y=x + 2\)
10. A: \(10\); B: \(-\frac{5}{3}\); C: \(y=-\frac{5}{3}x + 10\) (estimates may vary based on graph interpretation)