Why are polynomials tangential to the $x$ axis at real double roots?
$begingroup$
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
558112
$endgroup$
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
558112
$endgroup$
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
558112
$endgroup$
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
558112
asked Jan 16 at 16:46
GnumbertesterGnumbertester
558112
asked Jan 16 at 16:46
GnumbertesterGnumbertester
558112
asked Jan 16 at 16:46
GnumbertesterGnumbertester
558112
asked Jan 16 at 16:46
GnumbertesterGnumbertester
558112
558112
add a comment |
add a comment |
2 Answers
2
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
26.7k22157
$endgroup$
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
3,113417
$endgroup$
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075961%2fwhy-are-polynomials-tangential-to-the-x-axis-at-real-double-roots%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
26.7k22157
$endgroup$
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
26.7k22157
$endgroup$
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
26.7k22157
$endgroup$
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
26.7k22157
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.7k22157
answered Jan 16 at 16:55
mathcounterexamples.netmathcounterexamples.net
26.7k22157
26.7k22157
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
3,113417
$endgroup$
$begingroup$
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
3,113417
$endgroup$
$begingroup$
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
3,113417
$endgroup$
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
3,113417
answered Jan 16 at 16:50
pwerthpwerth
3,113417
answered Jan 16 at 16:50
pwerthpwerth
3,113417
answered Jan 16 at 16:50
pwerthpwerth
3,113417
3,113417
$begingroup$
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075961%2fwhy-are-polynomials-tangential-to-the-x-axis-at-real-double-roots%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
