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How to Find the Slope, Y-intercept, and Equation of a Line Given a Table of Values

Step 1 Pick any two points from the postpone, and write them as coordinate pairs .
Step 2: Pick one bespeak to be { eq } ( \text X_1, \text Y_1 ) { /eq }, and the early point will be { eq } ( \text X_2, \text Y_2 ) { /eq } .
Step 3: Substitute the values from these points into the slope formula, { equivalent } \dfrac { \text Y_1-\text Y_2 } { \text X_1-\text X_2 } { /eq }, and simplify .
Step 4: Pick one of the points you used to calculate the slope, and substitute the y and x values into the slope-intercept mannequin of a linear routine, { equivalent } y=mx+b { /eq } .
Step 5: Replace m with the slope you calculated in measure 3. Simplify both sides of the equation before moving on to the next step .
Step 6: Use inverse operations to solve the equality from measure 5 for b .
Step 7: Plug the values we found for b and m into the slope-intercept form of a linear affair appropriately. This gives you the linear function for the table of values. The variables x and y should remain letters in our concluding answer .

How to Find the Slope, Y-intercept, and Equation of a Line Given a Table of Values Vocabulary

Linear Function: This is a function that has a changeless rate of change and produces a uncoiled telephone line when graph .
Slope-Intercept Form of a Linear Function: This is a cosmopolitan form of a linear function given by the equation { equivalent } y=mx+b { /eq }. In this equality, m stands for the gradient of the function, and the b stands for the y-intercept of the function .
so, permit ‘s try using these steps and definitions to find the slope, y-intercept, and equation for a linear function given a values table in the following two examples .

How to Find the Slope, Y-intercept, and Equation of a Line Given a Table of Values: Example 1

What is the slope-intercept imprint of the linear function for the adopt postpone of values ?

x y
-2 10
-1 7
0 4
1 1
2 -2

Step 1 Pick any two points from the table, and write them as coordinate pairs .
We can use any two points, so lease ‘s say we use { equivalent } ( 1, 1 ) { /eq } and { equivalent } ( -2, 10 ) { /eq } .
Step 2: Pick one compass point to be { eq } ( \text X_1, \text Y_1 ) { /eq }, and the other item will be { eq } ( \text X_2, \text Y_2 ) { /eq } .

  • Let’s let {eq}(\text X_1, \text Y_1)

    {/eq} be {eq}(1,1)

    {/eq}.

  • Let’s let {eq}(\text X_2, \text Y_2)

    {/eq} be {eq}(-2,10)

    {/eq}.

Step 3: Substitute the values from these points into the gradient recipe, { equivalent } \dfrac { \text Y_1-\text Y_2 } { \text X_1-\text X_2 } { /eq }, and simplify .
We plug { equivalent } X_1=1, \ ; Y_1=1, \ ; X_2=-2, \ ; \text { and } \ ; Y_2=10 { /eq } into our slope formula and simplify as follows :
$ $ \begin { align } thousand & =\dfrac { \text Y_1-\text Y_2 } { \text X_1-\text X_2 } \\\\ & =\dfrac { 1-10 } { 1- ( -2 ) } \\\\ & =\dfrac { -9 } { 3 } \\\\ & =-3 \end { align } $ $
The slope of our line is m = -3.

Step 4: Pick one of the points you used to calculate gradient, and substitute the y and x values into the slope-intercept form of a linear function, { equivalent } y=mx+b { /eq } .
If we use our first base point ( 1,1 ), we plug x = 1 and y = 1 into our equality to get :
$ $ 1=m ( 1 ) +b $ $
It is helpful to write parentheses around the varying we are replacing if there is another variable multiplied by that measure .
Step 5: Replace m with the gradient you calculated in step 3. Simplify both sides of the equation before moving on to the following footfall .
now, we plug m = -3 into this equality to get the comply :
$ $ 1= ( -3 ) ( 1 ) +b $ $
We see that we put digression around the variable we are replacing in this step ampere well since there is multiplication. immediately, we simplify .
$ $ \begin { align } 1 & = ( -3 ) ( 1 ) +b\\ 1 & =-3+b \end { align } $ $
Step 6: Use inverse operations to solve the equation from step 5 for b .
Solving our equation for b goes as follows :
$ $ \begin { align } 1 & =-3+b\\ 1+3 & =-3+3+b\\ 4 & =b \end { align } $ $
Step 7: Plug the values we found for b and m into the slope-intercept form of a linear routine appropriately. This gives you the linear affair for the table of values. The variables x and y should remain letters in our final answer .
We found b = 4 and m = -3. Plugging these into the slope-intercept imprint of a linear officiate appropriately gives our equality as :
$ $ \begin { align } y & =mx+b\\ y & =-3x+4 \end { align } $ $
thus, our equation is { equivalent } y=-3x+4 { /eq } .

How to Find the Slope, Y-intercept, and Equation of a Line Given a Table of Values: Example 2

What is the slope-intercept kind of the linear equation for the surveil table of values ?

x y
3 1
6 3
9 5
12 7

Step 1: Choose our points .
{ equivalent } ( 3,1 ) { /eq } and { equivalent } ( 6,3 ) { /eq }
Step 2: Let one point be our first point and the early distributor point be our second point .

  • {eq}(\text X_1, \text Y_1)\rightarrow(3,1)

    {/eq}

  • {eq}(\text X_2, \text Y_2)\rightarrow(6,3)

    {/eq}

Step 3: Calculate our gradient .
$ $ \begin { align } \dfrac { 1-3 } { 3-6 } & =\dfrac { -2 } { -3 } \\\\ & =\dfrac { 2 } { 3 } \end { align } $ $
Step 4: Plug x and y into the slope-intercept form of a linear function .
$ $ 3=m ( 6 ) +b $ $
Step 5: Plug m into this equation, and simplify .
$ $ \begin { align } 3 & =\dfrac { 2 } { 3 } ( 6 ) +b\\\\ 3 & =\dfrac { 2 } { 3 } \cdot\dfrac { 6 } { 1 } +b\\\\ 3 & =\dfrac { 2\cdot6 } { 3\cdot1 } +b\\\\ 3 & =\dfrac { 12 } { 3 } +b\\\\ 3 & =4+b \end { align } $ $
Step 6: Solve for b .
$ $ \begin { align } 3 & =4+b\\ 3-4 & =4-4+b\\ -1 & =b \end { align } $ $

Step 7: Plug m and b into the slope-intercept form of a linear function .
$ $ y=\dfrac { 2 } { 3 } x-1 $ $

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