Quadratics: Intuitive relation between discriminant and derivative at roots
$begingroup$
While working with quadratics that have real roots, I realized an interesting fact:
The slope of a quadratic at its roots is equal to $pm sqrt{D}$ where $D=b^2-4ac$
Proof:
$$f(x) = ax^2 + bx +c$$
$$f'(x) = 2ax+b$$
Roots:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
So, if we try to find the slope at any root ($r$):
$$f’(r) = pm sqrt D$$
where the sign ($pm$) can be determined by whether the root is on the right of the vertex or the left.
If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.
What geometric/intuitive approach can be applied to explain this interesting phenomenon?
derivatives polynomials roots quadratics slope
edited Jan 14 at 11:34
KM101
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
$endgroup$
add a comment |
$begingroup$
While working with quadratics that have real roots, I realized an interesting fact:
The slope of a quadratic at its roots is equal to $pm sqrt{D}$ where $D=b^2-4ac$
Proof:
$$f(x) = ax^2 + bx +c$$
$$f'(x) = 2ax+b$$
Roots:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
So, if we try to find the slope at any root ($r$):
$$f’(r) = pm sqrt D$$
where the sign ($pm$) can be determined by whether the root is on the right of the vertex or the left.
If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.
What geometric/intuitive approach can be applied to explain this interesting phenomenon?
derivatives polynomials roots quadratics slope
edited Jan 14 at 11:34
KM101
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
$endgroup$
1
$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15
add a comment |
$begingroup$
While working with quadratics that have real roots, I realized an interesting fact:
The slope of a quadratic at its roots is equal to $pm sqrt{D}$ where $D=b^2-4ac$
Proof:
$$f(x) = ax^2 + bx +c$$
$$f'(x) = 2ax+b$$
Roots:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
So, if we try to find the slope at any root ($r$):
$$f’(r) = pm sqrt D$$
where the sign ($pm$) can be determined by whether the root is on the right of the vertex or the left.
If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.
What geometric/intuitive approach can be applied to explain this interesting phenomenon?
derivatives polynomials roots quadratics slope
edited Jan 14 at 11:34
KM101
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
$endgroup$
While working with quadratics that have real roots, I realized an interesting fact:
The slope of a quadratic at its roots is equal to $pm sqrt{D}$ where $D=b^2-4ac$
Proof:
$$f(x) = ax^2 + bx +c$$
$$f'(x) = 2ax+b$$
Roots:
$$x = frac{-bpmsqrt{b^2-4ac}}{2a}$$
So, if we try to find the slope at any root ($r$):
$$f’(r) = pm sqrt D$$
where the sign ($pm$) can be determined by whether the root is on the right of the vertex or the left.
If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.
What geometric/intuitive approach can be applied to explain this interesting phenomenon?
derivatives polynomials roots quadratics slope
derivatives polynomials roots quadratics slope
edited Jan 14 at 11:34
KM101
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
edited Jan 14 at 11:34
KM101
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
edited Jan 14 at 11:34
KM101
6,0861525
edited Jan 14 at 11:34
KM101
6,0861525
edited Jan 14 at 11:34
KM101
6,0861525
6,0861525
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
asked Jan 14 at 11:10
Vikhyat AgarwalVikhyat Agarwal
1163
1163
1
$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15
add a comment |
1
$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15
1
1
$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15
$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15
add a comment |
1 Answer
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I don’t see there is any special geometric related interpretation.
We will just discuss the case that $D gt 0$ such that $alpha$ and $beta$ are the two real roots with $alpha lt beta$.
For simplicity, we also drop the $pm$ sign.
$sqrt D = dfrac aa sqrt D$
$ = asqrt {(dfrac {-b}{a})^2 – dfrac {4c}{a}}$
$= a sqrt {(beta + alpha)^2 – 4alphabeta}$
$= a sqrt {(beta - alpha)^2}$
$= a times (beta - alpha)$
At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.
answered Jan 15 at 10:03
MickMick
12.1k31641
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add a comment |
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$begingroup$
I don’t see there is any special geometric related interpretation.
We will just discuss the case that $D gt 0$ such that $alpha$ and $beta$ are the two real roots with $alpha lt beta$.
For simplicity, we also drop the $pm$ sign.
$sqrt D = dfrac aa sqrt D$
$ = asqrt {(dfrac {-b}{a})^2 – dfrac {4c}{a}}$
$= a sqrt {(beta + alpha)^2 – 4alphabeta}$
$= a sqrt {(beta - alpha)^2}$
$= a times (beta - alpha)$
At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.
answered Jan 15 at 10:03
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
I don’t see there is any special geometric related interpretation.
We will just discuss the case that $D gt 0$ such that $alpha$ and $beta$ are the two real roots with $alpha lt beta$.
For simplicity, we also drop the $pm$ sign.
$sqrt D = dfrac aa sqrt D$
$ = asqrt {(dfrac {-b}{a})^2 – dfrac {4c}{a}}$
$= a sqrt {(beta + alpha)^2 – 4alphabeta}$
$= a sqrt {(beta - alpha)^2}$
$= a times (beta - alpha)$
At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.
answered Jan 15 at 10:03
MickMick
12.1k31641
$endgroup$
add a comment |
$begingroup$
I don’t see there is any special geometric related interpretation.
We will just discuss the case that $D gt 0$ such that $alpha$ and $beta$ are the two real roots with $alpha lt beta$.
For simplicity, we also drop the $pm$ sign.
$sqrt D = dfrac aa sqrt D$
$ = asqrt {(dfrac {-b}{a})^2 – dfrac {4c}{a}}$
$= a sqrt {(beta + alpha)^2 – 4alphabeta}$
$= a sqrt {(beta - alpha)^2}$
$= a times (beta - alpha)$
At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.
answered Jan 15 at 10:03
MickMick
12.1k31641
$endgroup$
I don’t see there is any special geometric related interpretation.
We will just discuss the case that $D gt 0$ such that $alpha$ and $beta$ are the two real roots with $alpha lt beta$.
For simplicity, we also drop the $pm$ sign.
$sqrt D = dfrac aa sqrt D$
$ = asqrt {(dfrac {-b}{a})^2 – dfrac {4c}{a}}$
$= a sqrt {(beta + alpha)^2 – 4alphabeta}$
$= a sqrt {(beta - alpha)^2}$
$= a times (beta - alpha)$
At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.
answered Jan 15 at 10:03
MickMick
12.1k31641
answered Jan 15 at 10:03
MickMick
12.1k31641
answered Jan 15 at 10:03
MickMick
12.1k31641
answered Jan 15 at 10:03
MickMick
12.1k31641
12.1k31641
add a comment |
add a comment |
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$begingroup$
Section 3 of this paper offers some geometric insights for the quadratic: nickalls.org/dick/papers/maths/quadratic2000.pdf
$endgroup$
– Andy Walls
Jan 15 at 12:15