QuestionFind the slope, $y$ -intercept, and equation of the line for points (5, -8), (10, -12), (15, -16), (20, -20).

Studdy Solution

STEP 1

Assumptions1. The given values are points on a line in a Cartesian coordinate system. . The relationship between $x$ and $y$ is linear.
3. We are asked to find the slope, $y$ -intercept, and equation of the line.

STEP 2

First, we need to find the slope of the line. The slope (m) is the change in $y$ divided by the change in $x$ .
$m = \frac{\Delta y}{\Delta x}$

STEP 3

We can calculate the slope using any two points on the line. Let's use the points (5, -8) and (10, -12).
$m = \frac{-12 - (-8)}{10 -5}$

STEP 4

Calculate the slope.
$m = \frac{-12 - (-8)}{10 -} = \frac{-4}{} = -0.8$

STEP 5

Now that we have the slope, we can find the $y$ -intercept. The $y$ -intercept (b) is the $y$ value when $x$ is0. We can find this by rearranging the equation of a line, $y = mx + b$ , to solve for $b$ .
$b = y - mx$

STEP 6

We can use any point on the line to find the $y$ -intercept. Let's use the point (5, -8).
$b = -8 - (-0.8 \times5)$

STEP 7

Calculate the $y$ -intercept.
$b = - - (-0. \times5) = - +4 = -4$

STEP 8

Now that we have the slope and the $y$ -intercept, we can write the equation of the line. The equation of a line is given by $y = mx + b$ .
$y = -0.8x -4$ The slope of the line is -0.8, the $y$ -intercept is -4, and the equation of the line is $y = -0.8x -4$ .

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QuestionFind the slope, $y$ -intercept, and equation of the line for points (5, -8), (10, -12), (15, -16), (20, -20).

Studdy Solution

STEP 1

Assumptions1. The given values are points on a line in a Cartesian coordinate system. . The relationship between $x$ and $y$ is linear.
3. We are asked to find the slope, $y$ -intercept, and equation of the line.

STEP 2

First, we need to find the slope of the line. The slope (m) is the change in $y$ divided by the change in $x$ .
$m = \frac{\Delta y}{\Delta x}$

STEP 3

We can calculate the slope using any two points on the line. Let's use the points (5, -8) and (10, -12).
$m = \frac{-12 - (-8)}{10 -5}$

STEP 4

Calculate the slope.
$m = \frac{-12 - (-8)}{10 -} = \frac{-4}{} = -0.8$

STEP 5

Now that we have the slope, we can find the $y$ -intercept. The $y$ -intercept (b) is the $y$ value when $x$ is0. We can find this by rearranging the equation of a line, $y = mx + b$ , to solve for $b$ .
$b = y - mx$

STEP 6

We can use any point on the line to find the $y$ -intercept. Let's use the point (5, -8).
$b = -8 - (-0.8 \times5)$

STEP 7

Calculate the $y$ -intercept.
$b = - - (-0. \times5) = - +4 = -4$

STEP 8

Now that we have the slope and the $y$ -intercept, we can write the equation of the line. The equation of a line is given by $y = mx + b$ .
$y = -0.8x -4$ The slope of the line is -0.8, the $y$ -intercept is -4, and the equation of the line is $y = -0.8x -4$ .

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QuestionFind the slope, yyy-intercept, and equation of the line for points (5, -8), (10, -12), (15, -16), (20, -20).

Studdy Solution

STEP 1

Assumptions1. The given values are points on a line in a Cartesian coordinate system. . The relationship between xxx and yyy is linear.3. We are asked to find the slope, yyy-intercept, and equation of the line.

STEP 2

First, we need to find the slope of the line. The slope (m) is the change in yyy divided by the change in xxx.m=ΔyΔxm = \frac{\Delta y}{\Delta x}m=ΔxΔy​

STEP 3

We can calculate the slope using any two points on the line. Let's use the points (5, -8) and (10, -12).m=−12−(−8)10−5m = \frac{-12 - (-8)}{10 -5}m=10−5−12−(−8)​

STEP 4

Calculate the slope.m=−12−(−8)10−=−4=−0.8m = \frac{-12 - (-8)}{10 -} = \frac{-4}{} = -0.8m=10−−12−(−8)​=−4​=−0.8

STEP 5

Now that we have the slope, we can find the yyy-intercept. The yyy-intercept (b) is the yyy value when xxx is0. We can find this by rearranging the equation of a line, y=mx+by = mx + by=mx+b, to solve for bbb.b=y−mxb = y - mxb=y−mx

STEP 6

We can use any point on the line to find the yyy-intercept. Let's use the point (5, -8).b=−8−(−0.8×5)b = -8 - (-0.8 \times5)b=−8−(−0.8×5)

STEP 7

Calculate the yyy-intercept.b=−−(−0.×5)=−+4=−4b = - - (-0. \times5) = - +4 = -4b=−−(−0.×5)=−+4=−4

STEP 8

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QuestionFind the slope, $y$ -intercept, and equation of the line for points (5, -8), (10, -12), (15, -16), (20, -20).

Assumptions1. The given values are points on a line in a Cartesian coordinate system. . The relationship between $x$ and $y$ is linear.
3. We are asked to find the slope, $y$ -intercept, and equation of the line.

First, we need to find the slope of the line. The slope (m) is the change in $y$ divided by the change in $x$ .
$m = \frac{\Delta y}{\Delta x}$

We can calculate the slope using any two points on the line. Let's use the points (5, -8) and (10, -12).
$m = \frac{-12 - (-8)}{10 -5}$

Calculate the slope.
$m = \frac{-12 - (-8)}{10 -} = \frac{-4}{} = -0.8$

Now that we have the slope, we can find the $y$ -intercept. The $y$ -intercept (b) is the $y$ value when $x$ is0. We can find this by rearranging the equation of a line, $y = mx + b$ , to solve for $b$ .
$b = y - mx$

We can use any point on the line to find the $y$ -intercept. Let's use the point (5, -8).
$b = -8 - (-0.8 \times5)$

Calculate the $y$ -intercept.
$b = - - (-0. \times5) = - +4 = -4$