Mixing
Determine where peak or valley of polynomial graph without calculus
$begingroup$
I notice that if I graph $y=(x+2)^2(x-4)^2$ that the midpoint of the roots occurs at $x=1$. I note that this is also where the local maximum occurs.
If I graph $y=(x-5)^2(x^2)$ the midpoint of the roots is 2.5 where the local maximum occurs.
I also know how to find the exact values of the local max / min by taking the derivative and setting to zero.
My question is, for those in pre-calculus who have not yet learned about derivatives, is there a relationship between where the roots occur and the local max / min values? Certainly for the 2 examples I chose, there seems to be a relationship, but I can't see the mathematical reasoning for this. Calculus explains this, but could I predict the local max/min by simply finding the mid-point of the roots?
algebra-precalculus
asked Jan 18 at 21:21
user163862user163862
88021017
$endgroup$
add a comment |
$begingroup$
I notice that if I graph $y=(x+2)^2(x-4)^2$ that the midpoint of the roots occurs at $x=1$. I note that this is also where the local maximum occurs.
If I graph $y=(x-5)^2(x^2)$ the midpoint of the roots is 2.5 where the local maximum occurs.
I also know how to find the exact values of the local max / min by taking the derivative and setting to zero.
My question is, for those in pre-calculus who have not yet learned about derivatives, is there a relationship between where the roots occur and the local max / min values? Certainly for the 2 examples I chose, there seems to be a relationship, but I can't see the mathematical reasoning for this. Calculus explains this, but could I predict the local max/min by simply finding the mid-point of the roots?
algebra-precalculus
asked Jan 18 at 21:21
user163862user163862
88021017
$endgroup$
$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39
add a comment |
$begingroup$
I notice that if I graph $y=(x+2)^2(x-4)^2$ that the midpoint of the roots occurs at $x=1$. I note that this is also where the local maximum occurs.
If I graph $y=(x-5)^2(x^2)$ the midpoint of the roots is 2.5 where the local maximum occurs.
I also know how to find the exact values of the local max / min by taking the derivative and setting to zero.
My question is, for those in pre-calculus who have not yet learned about derivatives, is there a relationship between where the roots occur and the local max / min values? Certainly for the 2 examples I chose, there seems to be a relationship, but I can't see the mathematical reasoning for this. Calculus explains this, but could I predict the local max/min by simply finding the mid-point of the roots?
algebra-precalculus
asked Jan 18 at 21:21
user163862user163862
88021017
$endgroup$
I notice that if I graph $y=(x+2)^2(x-4)^2$ that the midpoint of the roots occurs at $x=1$. I note that this is also where the local maximum occurs.
If I graph $y=(x-5)^2(x^2)$ the midpoint of the roots is 2.5 where the local maximum occurs.
I also know how to find the exact values of the local max / min by taking the derivative and setting to zero.
My question is, for those in pre-calculus who have not yet learned about derivatives, is there a relationship between where the roots occur and the local max / min values? Certainly for the 2 examples I chose, there seems to be a relationship, but I can't see the mathematical reasoning for this. Calculus explains this, but could I predict the local max/min by simply finding the mid-point of the roots?
algebra-precalculus
algebra-precalculus
asked Jan 18 at 21:21
user163862user163862
88021017
asked Jan 18 at 21:21
user163862user163862
88021017
asked Jan 18 at 21:21
user163862user163862
88021017
asked Jan 18 at 21:21
user163862user163862
88021017
asked Jan 18 at 21:21
user163862user163862
88021017
88021017
$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39
add a comment |
$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39
$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
add a comment |
$begingroup$
There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
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$begingroup$
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
add a comment |
$begingroup$
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
$endgroup$
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
add a comment |
$begingroup$
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
$endgroup$
You can start with a parabola. If it has two real roots we can write it in the form $a(x-b)(x-c)$ with the roots at $b,c$. The factor $a$ does not change the location of the roots or the vertex. By using the quadratic formula or calculus you can find that the vertex is at $x=frac 12(b+c)$, the midpoint between the roots. Whether the vertex is above or below the axis, when you square this quadratic the local minima will be $0$ at the original roots and the local maximum will be at the location of the vertex, which is halfway between the roots.
If you have a quartic with two double roots it must be of the form $a(x-b)^2(x-c)^2$ and what you have noticed is correct. If you look at more general quartics you can have different behavior. If we consider $xy=x^2(x-1)^2+frac x{10}$ we now have roots at $0$ and about $-0.085$ but the local maximum is near $x=0.6$. A graph is below.
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
answered Jan 18 at 21:43
Ross MillikanRoss Millikan
298k23198371
298k23198371
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
add a comment |
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
$begingroup$
Thank you. Very clear now.
$endgroup$
– user163862
Jan 18 at 22:38
add a comment |
$begingroup$
There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
$endgroup$
add a comment |
$begingroup$
There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
$endgroup$
There is no precise relation because you can keep the extrema fixed while the roots are moving. Consider the cubic
$$x^3-3x+c,$$ that always has a maximum at $x=-1$ and a minimum at $x=1$.
We can place a root wherever we want, say at $x_0$, just by setting
$$c=-x_0^3+3x_0.$$
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
answered Jan 18 at 21:32
Yves DaoustYves Daoust
129k675227
129k675227
add a comment |
add a comment |
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$begingroup$
Are you specifically asking about these functions of the form $(x+a)^2 (x+b)^2$ or about all polynomials?
$endgroup$
– Michael Lugo
Jan 18 at 21:27
$begingroup$
Not necessarily, if you're considering all polynomials in one variable. Consider $y=x^3$. It's root is $0$ (multiplicty 3). $x=0$ is the point of inflection is at $x=0$, and it has no max or min.
$endgroup$
– jordan_glen
Jan 18 at 21:27
$begingroup$
I was asking about polynomials with real roots. In the precalc book there were several of the form $(x+a)^2(x+b)^2$ and I noticed that the midpoint of the segment between the roots corresponded to the "local" max/min value.
$endgroup$
– user163862
Jan 18 at 21:39