Finding a linear function from given functions
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
algebra-precalculus functions polynomials
edited Dec 27 at 15:30
greedoid
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
add a comment |
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
algebra-precalculus functions polynomials
edited Dec 27 at 15:30
greedoid
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29
add a comment |
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
algebra-precalculus functions polynomials
edited Dec 27 at 15:30
greedoid
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
The question is asking to find the linear function, $f(t) = vt + C$ for $f(t+2) = f(t) + 6$ and $f(1) = 10$
The answer is $3t + 7$, but I have no idea how the answer is produced.
algebra-precalculus functions polynomials
algebra-precalculus functions polynomials
edited Dec 27 at 15:30
greedoid
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
edited Dec 27 at 15:30
greedoid
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
edited Dec 27 at 15:30
greedoid
37.7k114794
edited Dec 27 at 15:30
greedoid
37.7k114794
edited Dec 27 at 15:30
greedoid
37.7k114794
37.7k114794
asked Dec 27 at 15:25
Evan Kim
998
asked Dec 27 at 15:25
Evan Kim
998
asked Dec 27 at 15:25
Evan Kim
998
998
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29
add a comment |
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29
Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29
add a comment |
3 Answers
3
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Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 at 15:28
greedoid
37.7k114794
add a comment |
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 at 15:29
KM101
4,836421
add a comment |
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 at 22:23
Eric Towers
31.8k22265
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 at 15:28
greedoid
37.7k114794
add a comment |
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 at 15:28
greedoid
37.7k114794
add a comment |
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 at 15:28
greedoid
37.7k114794
Since $f(1)=10$ we have $f(3)= f(1)+6=16$. Thus you have to solve the system:
$$ v+c=10$$
$$ 3v+c = 16$$
answered Dec 27 at 15:28
greedoid
37.7k114794
answered Dec 27 at 15:28
greedoid
37.7k114794
answered Dec 27 at 15:28
greedoid
37.7k114794
answered Dec 27 at 15:28
greedoid
37.7k114794
37.7k114794
add a comment |
add a comment |
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 at 15:29
KM101
4,836421
add a comment |
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 at 15:29
KM101
4,836421
add a comment |
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 at 15:29
KM101
4,836421
Hint:
$$f(t+2) = f(t)+6$$
$$f(1) = 10 = f(-1)+6 iff f(-1) = 4$$
You now have the two points $(1, 10)$ and $(-1, 4)$.
Small Addition: From here, you can either use the use a system of equations for the two points $(1, 10)$ and $(-1, 4)$:
$$begin{cases}
v+C = 10\
-v+C = 4
end{cases}$$
or you can refer to the definition of a linear equation $y = vt+C$, in which $v$ is the slope and $C$ is the $y$-intercept, or the $y$-coordinate when $x = 0$ (and it’s easy to find here because the point is the midpoint of the two points already found).
$$v = frac{Delta f(t)}{Delta t}$$
answered Dec 27 at 15:29
KM101
4,836421
edited Dec 27 at 16:51
answered Dec 27 at 15:29
KM101
4,836421
answered Dec 27 at 15:29
KM101
4,836421
answered Dec 27 at 15:29
KM101
4,836421
4,836421
add a comment |
add a comment |
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 at 22:23
Eric Towers
31.8k22265
add a comment |
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 at 22:23
Eric Towers
31.8k22265
add a comment |
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 at 22:23
Eric Towers
31.8k22265
... and now the conceptual version. "$f(t+2) = f(t)+6$" means that the slope is $6$ (the increase in output) over a run of $2$ (the increase in input), so the slope is $frac{6}{2} = 3$. Then you have a point on the line, $(1,10)$ from "$f(1) = 10$", so $f(t) - 10 = 3(t-1)$, by point-slope, and we have $f(t) = 3t + 7$.
answered Dec 27 at 22:23
Eric Towers
31.8k22265
answered Dec 27 at 22:23
Eric Towers
31.8k22265
answered Dec 27 at 22:23
Eric Towers
31.8k22265
answered Dec 27 at 22:23
Eric Towers
31.8k22265
31.8k22265
add a comment |
add a comment |
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Set up the two given conditions as equations to be solved for the coefficients $v$ and $C$.
– zipirovich
Dec 27 at 15:29