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Why two formulas for a in a linear function are equal
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Sorry if this is an easy question for some, but I have tried and failed for a while now, so I need someone else to help me.
I want to figure out why $$a=fracf(x)-bx_1$$ is equal to $$a=fracy_2-y_1x_2-x_1$$
For example in a task where you know to points in a linear graph, you use the second formula to find $a$.
$$A: (2, 10)\
B: (8, 130)$$
Then $a=20$.
But why is the flipped version of $f(x)=ax+b$ equal to the direct formula $a=fracy_2-y_1x_2-x_1$? Are they even the same? What I mean is that can the two formulas be flipped/multiplied and stuff like that to be the same?
linear-algebra
edited 1 hour ago
5xum
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
2
down vote
favorite
Sorry if this is an easy question for some, but I have tried and failed for a while now, so I need someone else to help me.
I want to figure out why $$a=fracf(x)-bx_1$$ is equal to $$a=fracy_2-y_1x_2-x_1$$
For example in a task where you know to points in a linear graph, you use the second formula to find $a$.
$$A: (2, 10)\
B: (8, 130)$$
Then $a=20$.
But why is the flipped version of $f(x)=ax+b$ equal to the direct formula $a=fracy_2-y_1x_2-x_1$? Are they even the same? What I mean is that can the two formulas be flipped/multiplied and stuff like that to be the same?
linear-algebra
edited 1 hour ago
5xum
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
1 hour ago
Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
1 hour ago
@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
– F.Carette
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Sorry if this is an easy question for some, but I have tried and failed for a while now, so I need someone else to help me.
I want to figure out why $$a=fracf(x)-bx_1$$ is equal to $$a=fracy_2-y_1x_2-x_1$$
For example in a task where you know to points in a linear graph, you use the second formula to find $a$.
$$A: (2, 10)\
B: (8, 130)$$
Then $a=20$.
But why is the flipped version of $f(x)=ax+b$ equal to the direct formula $a=fracy_2-y_1x_2-x_1$? Are they even the same? What I mean is that can the two formulas be flipped/multiplied and stuff like that to be the same?
linear-algebra
edited 1 hour ago
5xum
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sorry if this is an easy question for some, but I have tried and failed for a while now, so I need someone else to help me.
I want to figure out why $$a=fracf(x)-bx_1$$ is equal to $$a=fracy_2-y_1x_2-x_1$$
For example in a task where you know to points in a linear graph, you use the second formula to find $a$.
$$A: (2, 10)\
B: (8, 130)$$
Then $a=20$.
But why is the flipped version of $f(x)=ax+b$ equal to the direct formula $a=fracy_2-y_1x_2-x_1$? Are they even the same? What I mean is that can the two formulas be flipped/multiplied and stuff like that to be the same?
linear-algebra
linear-algebra
edited 1 hour ago
5xum
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
5xum
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
5xum
85.4k388154
edited 1 hour ago
5xum
85.4k388154
edited 1 hour ago
5xum
85.4k388154
85.4k388154
asked 1 hour ago
Adrian Fagerland
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Adrian Fagerland
111
asked 1 hour ago
Adrian Fagerland
111
111
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Adrian Fagerland is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
1 hour ago
Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
1 hour ago
@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
– F.Carette
1 hour ago
add a comment |Â
Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
1 hour ago
Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
1 hour ago
@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
– F.Carette
1 hour ago
Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
1 hour ago
Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
1 hour ago
Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
1 hour ago
Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
1 hour ago
@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
– F.Carette
1 hour ago
@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
– F.Carette
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
They are not equal. Your two equations feature different variables, so clearly, they cannot be the same. However, they are somewhat related, as both are closely linked to linear functions.
Short answer:
The first formula is used to calculate $a$ if you do know $b$ and you know one point on the graph of the function.
The second formula is used to calculate $a$ if you don't know $b$ and you know two points on the graph of the function.
Longer answer:
Both equations can be seen to represent the same thing, which is a linear function of $x$.
Remember: By definition, a linear function is a function of $x$ equal to $$f(x)=ax+b$$
From this equation, we can see that, by moving $b$ to the left side of the equation and dividing by $x$, the equation is equivalent to the equation $$a=fracf(x)-bx$$ so long as $xneq 0$.
Therefore, the above expression is an expression that is true for every point $(x,f(x))$ on the graph of the linear function.
I can use this formula to calculate $a$, for example, if I already know the value of $b$ and one single point on the line, and want to calculate $a$.
On the other hand, let's we take two points, $(x_1,y_1)$ and $(x_2,y_2)$. Then, let's imagine that the two points are on the same line, then $y_1=ax_1+b$ and $y_2=ax_2+b$. Subtracting these two equations, and some division later, we get the equation $$a=fracy_2-y_1x_2-x_1$$
which is an equation that is true for every pair of points on the graph of the linear function.
I can use this formula to calculate $a$ if I don't know what $b$ is, so long as I know two points on the graph.
Edit:
There is also a "third", special option for calculating $a$. Let's say you happen to know two points $(x_1, y_1)$ and $(x_2, y_2)$, and the $x$ value for one of them is $0$, that is, $x_1=0$. There are now two ways of calculating $a$.
Using the fact that $f(x)=ax+b$, you can plug in $x_1=0$ and get $y_1=acdot 0 + b$ which means that $y_1=b$. You now know the value of $b$ and one point on the equation, which is $(x_2,y_2)$, and you can use the first formula to calculate $a$ to get $$a=fracf(x_2)-bx_2\ a=fracy_2-y_1x_2\
a=fracy_2-y_1x_2-0\
a=fracy_2-y_1x_2-x_1$$
You can see that the last formula in this case is exactly the same as the second formula for calculating $a$ in your question.
answered 1 hour ago
5xum
85.4k388154
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
add a comment |Â
up vote
2
down vote
What is meant by all this is that if You consider an affine-linear function ( actually just the first part $xmapsto ax$ is linear but by abuse of language one calls the affine function $f$ linear too):
$$f(x)=ax+b$$
You can compute $a$ by computing for any $x_0neq 0$:
$$a=fracf(x_0)-bx_0$$.
On the other hand You know that given two points $A(x_1|y_1)$ and $B(x_2|y_2)$ and $x_1neq x_2$ there is exactly one affine-linear function $f$ thats graph passes through these points and since the slope of an affine-linear function is everywhere the same You can compute it by the difference-quotient:
$$a=fracy_2-y_1x_2-x_1.$$
And if the above function $f$ is this one affine-linear function that has a graph passing through $A$ and $B$ then of course You have:
$$fracy_2-y_1x_2-x_1=fracf(x_0)-bx_0$$
for any $x_0neq 0.$
answered 1 hour ago
Peter Melech
2,198811
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
They are not equal. Your two equations feature different variables, so clearly, they cannot be the same. However, they are somewhat related, as both are closely linked to linear functions.
Short answer:
The first formula is used to calculate $a$ if you do know $b$ and you know one point on the graph of the function.
The second formula is used to calculate $a$ if you don't know $b$ and you know two points on the graph of the function.
Longer answer:
Both equations can be seen to represent the same thing, which is a linear function of $x$.
Remember: By definition, a linear function is a function of $x$ equal to $$f(x)=ax+b$$
From this equation, we can see that, by moving $b$ to the left side of the equation and dividing by $x$, the equation is equivalent to the equation $$a=fracf(x)-bx$$ so long as $xneq 0$.
Therefore, the above expression is an expression that is true for every point $(x,f(x))$ on the graph of the linear function.
I can use this formula to calculate $a$, for example, if I already know the value of $b$ and one single point on the line, and want to calculate $a$.
On the other hand, let's we take two points, $(x_1,y_1)$ and $(x_2,y_2)$. Then, let's imagine that the two points are on the same line, then $y_1=ax_1+b$ and $y_2=ax_2+b$. Subtracting these two equations, and some division later, we get the equation $$a=fracy_2-y_1x_2-x_1$$
which is an equation that is true for every pair of points on the graph of the linear function.
I can use this formula to calculate $a$ if I don't know what $b$ is, so long as I know two points on the graph.
Edit:
There is also a "third", special option for calculating $a$. Let's say you happen to know two points $(x_1, y_1)$ and $(x_2, y_2)$, and the $x$ value for one of them is $0$, that is, $x_1=0$. There are now two ways of calculating $a$.
Using the fact that $f(x)=ax+b$, you can plug in $x_1=0$ and get $y_1=acdot 0 + b$ which means that $y_1=b$. You now know the value of $b$ and one point on the equation, which is $(x_2,y_2)$, and you can use the first formula to calculate $a$ to get $$a=fracf(x_2)-bx_2\ a=fracy_2-y_1x_2\
a=fracy_2-y_1x_2-0\
a=fracy_2-y_1x_2-x_1$$
You can see that the last formula in this case is exactly the same as the second formula for calculating $a$ in your question.
answered 1 hour ago
5xum
85.4k388154
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
add a comment |Â
up vote
2
down vote
They are not equal. Your two equations feature different variables, so clearly, they cannot be the same. However, they are somewhat related, as both are closely linked to linear functions.
Short answer:
The first formula is used to calculate $a$ if you do know $b$ and you know one point on the graph of the function.
The second formula is used to calculate $a$ if you don't know $b$ and you know two points on the graph of the function.
Longer answer:
Both equations can be seen to represent the same thing, which is a linear function of $x$.
Remember: By definition, a linear function is a function of $x$ equal to $$f(x)=ax+b$$
From this equation, we can see that, by moving $b$ to the left side of the equation and dividing by $x$, the equation is equivalent to the equation $$a=fracf(x)-bx$$ so long as $xneq 0$.
Therefore, the above expression is an expression that is true for every point $(x,f(x))$ on the graph of the linear function.
I can use this formula to calculate $a$, for example, if I already know the value of $b$ and one single point on the line, and want to calculate $a$.
On the other hand, let's we take two points, $(x_1,y_1)$ and $(x_2,y_2)$. Then, let's imagine that the two points are on the same line, then $y_1=ax_1+b$ and $y_2=ax_2+b$. Subtracting these two equations, and some division later, we get the equation $$a=fracy_2-y_1x_2-x_1$$
which is an equation that is true for every pair of points on the graph of the linear function.
I can use this formula to calculate $a$ if I don't know what $b$ is, so long as I know two points on the graph.
Edit:
There is also a "third", special option for calculating $a$. Let's say you happen to know two points $(x_1, y_1)$ and $(x_2, y_2)$, and the $x$ value for one of them is $0$, that is, $x_1=0$. There are now two ways of calculating $a$.
Using the fact that $f(x)=ax+b$, you can plug in $x_1=0$ and get $y_1=acdot 0 + b$ which means that $y_1=b$. You now know the value of $b$ and one point on the equation, which is $(x_2,y_2)$, and you can use the first formula to calculate $a$ to get $$a=fracf(x_2)-bx_2\ a=fracy_2-y_1x_2\
a=fracy_2-y_1x_2-0\
a=fracy_2-y_1x_2-x_1$$
You can see that the last formula in this case is exactly the same as the second formula for calculating $a$ in your question.
answered 1 hour ago
5xum
85.4k388154
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
They are not equal. Your two equations feature different variables, so clearly, they cannot be the same. However, they are somewhat related, as both are closely linked to linear functions.
Short answer:
The first formula is used to calculate $a$ if you do know $b$ and you know one point on the graph of the function.
The second formula is used to calculate $a$ if you don't know $b$ and you know two points on the graph of the function.
Longer answer:
Both equations can be seen to represent the same thing, which is a linear function of $x$.
Remember: By definition, a linear function is a function of $x$ equal to $$f(x)=ax+b$$
From this equation, we can see that, by moving $b$ to the left side of the equation and dividing by $x$, the equation is equivalent to the equation $$a=fracf(x)-bx$$ so long as $xneq 0$.
Therefore, the above expression is an expression that is true for every point $(x,f(x))$ on the graph of the linear function.
I can use this formula to calculate $a$, for example, if I already know the value of $b$ and one single point on the line, and want to calculate $a$.
On the other hand, let's we take two points, $(x_1,y_1)$ and $(x_2,y_2)$. Then, let's imagine that the two points are on the same line, then $y_1=ax_1+b$ and $y_2=ax_2+b$. Subtracting these two equations, and some division later, we get the equation $$a=fracy_2-y_1x_2-x_1$$
which is an equation that is true for every pair of points on the graph of the linear function.
I can use this formula to calculate $a$ if I don't know what $b$ is, so long as I know two points on the graph.
Edit:
There is also a "third", special option for calculating $a$. Let's say you happen to know two points $(x_1, y_1)$ and $(x_2, y_2)$, and the $x$ value for one of them is $0$, that is, $x_1=0$. There are now two ways of calculating $a$.
Using the fact that $f(x)=ax+b$, you can plug in $x_1=0$ and get $y_1=acdot 0 + b$ which means that $y_1=b$. You now know the value of $b$ and one point on the equation, which is $(x_2,y_2)$, and you can use the first formula to calculate $a$ to get $$a=fracf(x_2)-bx_2\ a=fracy_2-y_1x_2\
a=fracy_2-y_1x_2-0\
a=fracy_2-y_1x_2-x_1$$
You can see that the last formula in this case is exactly the same as the second formula for calculating $a$ in your question.
answered 1 hour ago
5xum
85.4k388154
They are not equal. Your two equations feature different variables, so clearly, they cannot be the same. However, they are somewhat related, as both are closely linked to linear functions.
Short answer:
The first formula is used to calculate $a$ if you do know $b$ and you know one point on the graph of the function.
The second formula is used to calculate $a$ if you don't know $b$ and you know two points on the graph of the function.
Longer answer:
Both equations can be seen to represent the same thing, which is a linear function of $x$.
Remember: By definition, a linear function is a function of $x$ equal to $$f(x)=ax+b$$
From this equation, we can see that, by moving $b$ to the left side of the equation and dividing by $x$, the equation is equivalent to the equation $$a=fracf(x)-bx$$ so long as $xneq 0$.
Therefore, the above expression is an expression that is true for every point $(x,f(x))$ on the graph of the linear function.
I can use this formula to calculate $a$, for example, if I already know the value of $b$ and one single point on the line, and want to calculate $a$.
On the other hand, let's we take two points, $(x_1,y_1)$ and $(x_2,y_2)$. Then, let's imagine that the two points are on the same line, then $y_1=ax_1+b$ and $y_2=ax_2+b$. Subtracting these two equations, and some division later, we get the equation $$a=fracy_2-y_1x_2-x_1$$
which is an equation that is true for every pair of points on the graph of the linear function.
I can use this formula to calculate $a$ if I don't know what $b$ is, so long as I know two points on the graph.
Edit:
There is also a "third", special option for calculating $a$. Let's say you happen to know two points $(x_1, y_1)$ and $(x_2, y_2)$, and the $x$ value for one of them is $0$, that is, $x_1=0$. There are now two ways of calculating $a$.
Using the fact that $f(x)=ax+b$, you can plug in $x_1=0$ and get $y_1=acdot 0 + b$ which means that $y_1=b$. You now know the value of $b$ and one point on the equation, which is $(x_2,y_2)$, and you can use the first formula to calculate $a$ to get $$a=fracf(x_2)-bx_2\ a=fracy_2-y_1x_2\
a=fracy_2-y_1x_2-0\
a=fracy_2-y_1x_2-x_1$$
You can see that the last formula in this case is exactly the same as the second formula for calculating $a$ in your question.
answered 1 hour ago
5xum
85.4k388154
edited 1 hour ago
answered 1 hour ago
5xum
85.4k388154
answered 1 hour ago
5xum
85.4k388154
answered 1 hour ago
5xum
85.4k388154
85.4k388154
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
add a comment |Â
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
Just to nitpink a bit: you have $f(0)=b$. So, knowing $b$ is the same as knowing that $(0,b)$ belongs to the graph. You can then go back to the situation where you know two points of the graph. That's where the 2 equations are linked
– F.Carette
1 hour ago
1
1
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
@F.Carette Thanks for the thought, I edited my answer quite a bit to explain your comment.
– 5xum
1 hour ago
add a comment |Â
up vote
2
down vote
What is meant by all this is that if You consider an affine-linear function ( actually just the first part $xmapsto ax$ is linear but by abuse of language one calls the affine function $f$ linear too):
$$f(x)=ax+b$$
You can compute $a$ by computing for any $x_0neq 0$:
$$a=fracf(x_0)-bx_0$$.
On the other hand You know that given two points $A(x_1|y_1)$ and $B(x_2|y_2)$ and $x_1neq x_2$ there is exactly one affine-linear function $f$ thats graph passes through these points and since the slope of an affine-linear function is everywhere the same You can compute it by the difference-quotient:
$$a=fracy_2-y_1x_2-x_1.$$
And if the above function $f$ is this one affine-linear function that has a graph passing through $A$ and $B$ then of course You have:
$$fracy_2-y_1x_2-x_1=fracf(x_0)-bx_0$$
for any $x_0neq 0.$
answered 1 hour ago
Peter Melech
2,198811
add a comment |Â
up vote
2
down vote
What is meant by all this is that if You consider an affine-linear function ( actually just the first part $xmapsto ax$ is linear but by abuse of language one calls the affine function $f$ linear too):
$$f(x)=ax+b$$
You can compute $a$ by computing for any $x_0neq 0$:
$$a=fracf(x_0)-bx_0$$.
On the other hand You know that given two points $A(x_1|y_1)$ and $B(x_2|y_2)$ and $x_1neq x_2$ there is exactly one affine-linear function $f$ thats graph passes through these points and since the slope of an affine-linear function is everywhere the same You can compute it by the difference-quotient:
$$a=fracy_2-y_1x_2-x_1.$$
And if the above function $f$ is this one affine-linear function that has a graph passing through $A$ and $B$ then of course You have:
$$fracy_2-y_1x_2-x_1=fracf(x_0)-bx_0$$
for any $x_0neq 0.$
answered 1 hour ago
Peter Melech
2,198811
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What is meant by all this is that if You consider an affine-linear function ( actually just the first part $xmapsto ax$ is linear but by abuse of language one calls the affine function $f$ linear too):
$$f(x)=ax+b$$
You can compute $a$ by computing for any $x_0neq 0$:
$$a=fracf(x_0)-bx_0$$.
On the other hand You know that given two points $A(x_1|y_1)$ and $B(x_2|y_2)$ and $x_1neq x_2$ there is exactly one affine-linear function $f$ thats graph passes through these points and since the slope of an affine-linear function is everywhere the same You can compute it by the difference-quotient:
$$a=fracy_2-y_1x_2-x_1.$$
And if the above function $f$ is this one affine-linear function that has a graph passing through $A$ and $B$ then of course You have:
$$fracy_2-y_1x_2-x_1=fracf(x_0)-bx_0$$
for any $x_0neq 0.$
answered 1 hour ago
Peter Melech
2,198811
What is meant by all this is that if You consider an affine-linear function ( actually just the first part $xmapsto ax$ is linear but by abuse of language one calls the affine function $f$ linear too):
$$f(x)=ax+b$$
You can compute $a$ by computing for any $x_0neq 0$:
$$a=fracf(x_0)-bx_0$$.
On the other hand You know that given two points $A(x_1|y_1)$ and $B(x_2|y_2)$ and $x_1neq x_2$ there is exactly one affine-linear function $f$ thats graph passes through these points and since the slope of an affine-linear function is everywhere the same You can compute it by the difference-quotient:
$$a=fracy_2-y_1x_2-x_1.$$
And if the above function $f$ is this one affine-linear function that has a graph passing through $A$ and $B$ then of course You have:
$$fracy_2-y_1x_2-x_1=fracf(x_0)-bx_0$$
for any $x_0neq 0.$
answered 1 hour ago
Peter Melech
2,198811
edited 33 mins ago
answered 1 hour ago
Peter Melech
2,198811
answered 1 hour ago
Peter Melech
2,198811
answered 1 hour ago
Peter Melech
2,198811
2,198811
add a comment |Â
add a comment |Â
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Please read this before posting: math.meta.stackexchange.com/questions/5020/… to better format your questions in future. I fixed the formatting of this first question, but you should not expect others to do this for you!
– 5xum
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Thank you @5xum This is my first question like this, but I will do better in the future
– Adrian Fagerland
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@5xum Oups, failed the formatting of my link to the formatting page. Shame on me, and thank you for putting the correct link to the mathjax tutorial
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