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How to determine parameters $a, b,$ and $d$ so that a rational function models a given graph?
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I'm trying to solve a function problem. It states:
Determine the values of $a$, $b$, and $d$ so that the rational function $$f(x) = frac(x+a)(x-1)(x-b)(x-c)(x+d)(x-3)$$
correctly models this graph:
I've been looking at this for a while and I just can't figure out how I'm supposed to approach it. I think it has something to do with limits and I know they're all going to be integer numbers. Any ideas beyond randomly plugging numbers into Desmos until I get the right graph?
algebra-precalculus rational-functions
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
asked Aug 2 at 7:36
CaptainAmerica16
154110
add a comment |Â
up vote
5
down vote
favorite
I'm trying to solve a function problem. It states:
Determine the values of $a$, $b$, and $d$ so that the rational function $$f(x) = frac(x+a)(x-1)(x-b)(x-c)(x+d)(x-3)$$
correctly models this graph:
I've been looking at this for a while and I just can't figure out how I'm supposed to approach it. I think it has something to do with limits and I know they're all going to be integer numbers. Any ideas beyond randomly plugging numbers into Desmos until I get the right graph?
algebra-precalculus rational-functions
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
asked Aug 2 at 7:36
CaptainAmerica16
154110
6
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
2
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I'm trying to solve a function problem. It states:
Determine the values of $a$, $b$, and $d$ so that the rational function $$f(x) = frac(x+a)(x-1)(x-b)(x-c)(x+d)(x-3)$$
correctly models this graph:
I've been looking at this for a while and I just can't figure out how I'm supposed to approach it. I think it has something to do with limits and I know they're all going to be integer numbers. Any ideas beyond randomly plugging numbers into Desmos until I get the right graph?
algebra-precalculus rational-functions
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
asked Aug 2 at 7:36
CaptainAmerica16
154110
I'm trying to solve a function problem. It states:
Determine the values of $a$, $b$, and $d$ so that the rational function $$f(x) = frac(x+a)(x-1)(x-b)(x-c)(x+d)(x-3)$$
correctly models this graph:
I've been looking at this for a while and I just can't figure out how I'm supposed to approach it. I think it has something to do with limits and I know they're all going to be integer numbers. Any ideas beyond randomly plugging numbers into Desmos until I get the right graph?
algebra-precalculus rational-functions
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
asked Aug 2 at 7:36
CaptainAmerica16
154110
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
edited Aug 3 at 9:44
TheSimpliFire
9,40951751
9,40951751
asked Aug 2 at 7:36
CaptainAmerica16
154110
asked Aug 2 at 7:36
CaptainAmerica16
154110
asked Aug 2 at 7:36
CaptainAmerica16
154110
154110
6
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
2
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05
add a comment |Â
6
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
2
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05
6
6
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
2
2
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
13
down vote
accepted
from the expression, we see that whenever the numerator is zero, the graph intersects $x$ axis, and the numerator becomes zero when either of the factors is zero hence possible roots:
$x=-a, x=1$ and $x=b $. Also when the denominator is zero, the graph should tend to infinity if the numerator is non-zero at the same time. ie:
$x= c, x=-d$ and $x=3$.
From graph, roots are $x=-2,x=1$. Hence hints for $a$ and $b$.
Also as the discontinuity is at $x=3$, the function is not defined there hence the denominator is zero but the numerator is also but because nothing can be divided by zero, it is not defined at that single point. So you have to think of ways in which we can choose $c$ and $d$ such that at one of the function is not defined and at the other it simply reaches infinity.
You can take it from here, I guess.
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
add a comment |Â
up vote
7
down vote
HINT:
You can see that at $x=-1,2$, $f$ goes to infinity so $c$ and $d$ are ?
Then use the points $(-2,0)$ and $(1,0)$ to solve for $a$ and $b$.
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
add a comment |Â
up vote
3
down vote
For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc.
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$frac(0+2)(0-1)(0-b)(0-c)(0+d)(0-3).$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
from the expression, we see that whenever the numerator is zero, the graph intersects $x$ axis, and the numerator becomes zero when either of the factors is zero hence possible roots:
$x=-a, x=1$ and $x=b $. Also when the denominator is zero, the graph should tend to infinity if the numerator is non-zero at the same time. ie:
$x= c, x=-d$ and $x=3$.
From graph, roots are $x=-2,x=1$. Hence hints for $a$ and $b$.
Also as the discontinuity is at $x=3$, the function is not defined there hence the denominator is zero but the numerator is also but because nothing can be divided by zero, it is not defined at that single point. So you have to think of ways in which we can choose $c$ and $d$ such that at one of the function is not defined and at the other it simply reaches infinity.
You can take it from here, I guess.
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
add a comment |Â
up vote
13
down vote
accepted
from the expression, we see that whenever the numerator is zero, the graph intersects $x$ axis, and the numerator becomes zero when either of the factors is zero hence possible roots:
$x=-a, x=1$ and $x=b $. Also when the denominator is zero, the graph should tend to infinity if the numerator is non-zero at the same time. ie:
$x= c, x=-d$ and $x=3$.
From graph, roots are $x=-2,x=1$. Hence hints for $a$ and $b$.
Also as the discontinuity is at $x=3$, the function is not defined there hence the denominator is zero but the numerator is also but because nothing can be divided by zero, it is not defined at that single point. So you have to think of ways in which we can choose $c$ and $d$ such that at one of the function is not defined and at the other it simply reaches infinity.
You can take it from here, I guess.
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
from the expression, we see that whenever the numerator is zero, the graph intersects $x$ axis, and the numerator becomes zero when either of the factors is zero hence possible roots:
$x=-a, x=1$ and $x=b $. Also when the denominator is zero, the graph should tend to infinity if the numerator is non-zero at the same time. ie:
$x= c, x=-d$ and $x=3$.
From graph, roots are $x=-2,x=1$. Hence hints for $a$ and $b$.
Also as the discontinuity is at $x=3$, the function is not defined there hence the denominator is zero but the numerator is also but because nothing can be divided by zero, it is not defined at that single point. So you have to think of ways in which we can choose $c$ and $d$ such that at one of the function is not defined and at the other it simply reaches infinity.
You can take it from here, I guess.
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
from the expression, we see that whenever the numerator is zero, the graph intersects $x$ axis, and the numerator becomes zero when either of the factors is zero hence possible roots:
$x=-a, x=1$ and $x=b $. Also when the denominator is zero, the graph should tend to infinity if the numerator is non-zero at the same time. ie:
$x= c, x=-d$ and $x=3$.
From graph, roots are $x=-2,x=1$. Hence hints for $a$ and $b$.
Also as the discontinuity is at $x=3$, the function is not defined there hence the denominator is zero but the numerator is also but because nothing can be divided by zero, it is not defined at that single point. So you have to think of ways in which we can choose $c$ and $d$ such that at one of the function is not defined and at the other it simply reaches infinity.
You can take it from here, I guess.
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
edited Aug 3 at 9:45
TheSimpliFire
9,40951751
9,40951751
answered Aug 2 at 7:51
Zulu Raman
1484
answered Aug 2 at 7:51
Zulu Raman
1484
answered Aug 2 at 7:51
Zulu Raman
1484
1484
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
add a comment |Â
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
2
2
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
Welcome to the Mathematics Stack Exchange (Math.SE)! I must say, 'tis a beautiful answer :D
– user477343
Aug 3 at 9:47
add a comment |Â
up vote
7
down vote
HINT:
You can see that at $x=-1,2$, $f$ goes to infinity so $c$ and $d$ are ?
Then use the points $(-2,0)$ and $(1,0)$ to solve for $a$ and $b$.
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
add a comment |Â
up vote
7
down vote
HINT:
You can see that at $x=-1,2$, $f$ goes to infinity so $c$ and $d$ are ?
Then use the points $(-2,0)$ and $(1,0)$ to solve for $a$ and $b$.
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
add a comment |Â
up vote
7
down vote
up vote
7
down vote
HINT:
You can see that at $x=-1,2$, $f$ goes to infinity so $c$ and $d$ are ?
Then use the points $(-2,0)$ and $(1,0)$ to solve for $a$ and $b$.
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
HINT:
You can see that at $x=-1,2$, $f$ goes to infinity so $c$ and $d$ are ?
Then use the points $(-2,0)$ and $(1,0)$ to solve for $a$ and $b$.
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
answered Aug 2 at 7:42
TheSimpliFire
9,40951751
9,40951751
add a comment |Â
add a comment |Â
up vote
3
down vote
For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc.
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$frac(0+2)(0-1)(0-b)(0-c)(0+d)(0-3).$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
add a comment |Â
up vote
3
down vote
For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc.
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$frac(0+2)(0-1)(0-b)(0-c)(0+d)(0-3).$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
add a comment |Â
up vote
3
down vote
up vote
3
down vote
For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc.
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$frac(0+2)(0-1)(0-b)(0-c)(0+d)(0-3).$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
For this sort of problem, you should start by looking at "special points": intercepts, asymptotes, discontinuities, etc. You can then move on to looking for those for the derivative, second derivative, etc.
There are three special points highlighted for you: $x$-intercepts at $-2$ and $1$, and $y$-intercept at $1$, and a discontinuity at $(3,2.5)$. The $x$-intercepts tell you where the function should be zero: plugging in both $x = -2$ and $x = 1$ should give zero. If you plug $x = 1$ into $x - 1$, that gives you zero, so that's already taken care of. So now you need one of the parentheses to give you zero at $x = -2$. If $x = -2$, then adding $2$ to $x$ gives zero, so you want an $x+2$ in one of the parentheses, so $a = 2$.
For the $y$-intercept, you can plug in $x = 0$, and see that the equation gives you $$frac(0+2)(0-1)(0-b)(0-c)(0+d)(0-3).$$ This has three unknowns, so it's not very useful.
We can next look at the vertical asymptotes. There are two of them, at $x = -1$ and $x = 2$. A vertical asymptote corresponds to the denominator being zero, so we should have $x + 2$ and $x - 1$ on the bottom, giving $c = 1$ and $d = 2$. However, there's also an $x - 3$ on the bottom, but no vertical asymptote at $x = 3$. So we need to cancel that out with an $x - 3$ on the top, giving $b = 3$.
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
edited Aug 3 at 9:43
TheSimpliFire
9,40951751
9,40951751
answered Aug 2 at 19:08
Acccumulation
4,015314
answered Aug 2 at 19:08
Acccumulation
4,015314
answered Aug 2 at 19:08
Acccumulation
4,015314
4,015314
add a comment |Â
add a comment |Â
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6
You should immediately see that, according to the graph, the function is not defined at $x=-1$ and $x=2$ and at these points the graph goes to (plus or minus) infinity. Therefore, the resulting function should have those as the zeros of the denominator. Furthermore, the function is also not defined at $x=3$, as we can also see from the equation form. But this should "cancel out", as the function does not go to +/- infinity when approaching this point. These points should already get you pretty far ...
– Matti P.
Aug 2 at 7:41
2
... and the zeros of the function should correspond to zeros of the numerator. There are two of them on the graph, but the numerator has three zeros, so something should cancel out. The remaining consideration is the limits as x goes to $pminfty$.
– NickD
Aug 2 at 12:37
Thank you to everyone who provided help on this question. It not only helped me solve it but provided much-needed insight for the future!
– CaptainAmerica16
Aug 3 at 17:05