Why are polynomials tangential to the $x$ axis at real double roots?
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If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?
real-analysis calculus polynomials roots
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
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add a comment |
$begingroup$
If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?
real-analysis calculus polynomials roots
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
$endgroup$
add a comment |
$begingroup$
If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?
real-analysis calculus polynomials roots
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
$endgroup$
If $(x-a)^2$ is a root of a polynomial, then the graph will be tangent to the $x$ axis at $x=a$ but why? I know this is always the case for real double roots however I do not know the explanation for this behavior. Does this behavior apply to all real polynomial roots with even index multiplicities?
real-analysis calculus polynomials roots
real-analysis calculus polynomials roots
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
asked Jan 16 at 16:46
GnumbertesterGnumbertester
6771114
6771114
add a comment |
add a comment |
2 Answers
2
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First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^prime(x) = 2(x-a)q(x) + (x-a)^2 q^prime(x).$$
Therefore you have $p(a)=p^prime(a)=0$. Proving that the graph of the function $x mapsto p(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
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You can prove this easily using calculus, but I'll give an even more elementary argument.
The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).
Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.
Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:50
pwerthpwerth
3,310417
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$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^prime(x) = 2(x-a)q(x) + (x-a)^2 q^prime(x).$$
Therefore you have $p(a)=p^prime(a)=0$. Proving that the graph of the function $x mapsto p(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^prime(x) = 2(x-a)q(x) + (x-a)^2 q^prime(x).$$
Therefore you have $p(a)=p^prime(a)=0$. Proving that the graph of the function $x mapsto p(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^prime(x) = 2(x-a)q(x) + (x-a)^2 q^prime(x).$$
Therefore you have $p(a)=p^prime(a)=0$. Proving that the graph of the function $x mapsto p(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
First a comment
Saying that $(x-a)^2$ is a root of a polynomial $p(x)$ is an improper wording. You should say that $(x-a)^2$ divides $p(x)$ or that $a$ is a root of $p(x)$ of multiplicity at least equal to $2$.
Regarding your question
So you suppose that $p(x) = (x-a)^2q(x)$. Then taking the derivative you get
$$p^prime(x) = 2(x-a)q(x) + (x-a)^2 q^prime(x).$$
Therefore you have $p(a)=p^prime(a)=0$. Proving that the graph of the function $x mapsto p(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
$begingroup$
You can prove this easily using calculus, but I'll give an even more elementary argument.
The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).
Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.
Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:50
pwerthpwerth
3,310417
$endgroup$
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
add a comment |
$begingroup$
You can prove this easily using calculus, but I'll give an even more elementary argument.
The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).
Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.
Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:50
pwerthpwerth
3,310417
$endgroup$
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
add a comment |
$begingroup$
You can prove this easily using calculus, but I'll give an even more elementary argument.
The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).
Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.
Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:50
pwerthpwerth
3,310417
$endgroup$
You can prove this easily using calculus, but I'll give an even more elementary argument.
The graph of $f(x)=(x-a)^{2}$ is an upward facing parabola. In particular, it is the graph of $f(x)=x^{2}$ shifted to the right by $a$ units (if a is positive).
Note that, being a square, $(x-a)^{2}$ is always greater than or equal to $0$. And it is only equal to $0$ at $x=a$.
Combining the above two observations, the graph of $f(x)$ is tangent to the $x$-axis at $x=a$.
answered Jan 16 at 16:50
pwerthpwerth
3,310417
answered Jan 16 at 16:50
pwerthpwerth
3,310417
answered Jan 16 at 16:50
pwerthpwerth
3,310417
answered Jan 16 at 16:50
pwerthpwerth
3,310417
3,310417
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
add a comment |
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
$begingroup$
The OP is asking about any polynomial that has $(x-a)^2$ as a factor, not just $(x-a)^2$.
$endgroup$
– amd
Jan 16 at 18:24
add a comment |
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