Why does this quadratic expression have three zeroes?ODE with complex char roots gives strange...
Mathematica notebook opening off the screen
Optimization terminology: "Exact" v. "Approximate"
What is a solution?
Does throwing a penny at a train stop the train?
How can a dictatorship government be beneficial to a dictator in a post-scarcity society?
How to memorize multiple pieces?
How can I effectively communicate to recruiters that a phone call is not possible?
Did the Vulgar Latin verb "toccare" exist?
Why didn't Thanos kill all the Dwarves on Nidavellir?
Keep milk (or milk alternative) for a day without a fridge
Can fluent English speakers distinguish “steel”, “still” and “steal”?
Print the last, middle and first character of your code
Why does the U.S. tolerate foreign influence from Saudi Arabia and Israel on its domestic policies while not tolerating that from China or Russia?
Why can a destructor change the state of a constant object?
Is anyone advocating the promotion of homosexuality in UK schools?
How can I calculate the sum of 2 random dice out of a 3d6 pool in AnyDice?
Why do people keep referring to Leia as Princess Leia, even after the destruction of Alderaan?
Cracking the Coding Interview — 1.5 One Away
Why isn't there research to build a standard lunar, or Martian mobility platform?
How can one write good dialogue in a story without sounding wooden?
Why doesn't sea level show seasonality?
US Civil War story: man hanged from a bridge
Are neural networks prone to catastrophic forgetting?
Confirming the Identity of a (Friendly) Reviewer After the Reviews
Why does this quadratic expression have three zeroes?
ODE with complex char roots gives strange solutionsQuadratic equations: Why does factoring by grouping work?Find roots for an equation with quadratic, linear and log terms?Extracting factor from quadrinomialInfluence of small constant term on roots of polynomialtaking coefficient of highest order term zero in a polynomial and solving itQuestion related to a proof in Quadratic equationsPolynomial p(a)=1, why does it have at most 2 integer roots?Why are Quadratic progressions called so and how is general term of a quadratic sequence $an^2 + bn + c$?Roots of quadratic equation $ax^2+bx+c$ at infinity
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
$endgroup$
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
3
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago
add a comment |
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
$endgroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
roots quadratics
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
asked 9 hours ago
π times eπ times e
1492 silver badges9 bronze badges
1492 silver badges9 bronze badges
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
3
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago
add a comment |
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
3
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
3
3
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 9 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
Your Answer
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%2f3289009%2fwhy-does-this-quadratic-expression-have-three-zeroes%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 9 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
105k4 gold badges85 silver badges173 bronze badges
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 9 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 9 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 9 hours ago
DavidDavid
77410 bronze badges
$endgroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 9 hours ago
DavidDavid
77410 bronze badges
edited 9 hours ago
answered 9 hours ago
DavidDavid
77410 bronze badges
answered 9 hours ago
DavidDavid
77410 bronze badges
answered 9 hours ago
DavidDavid
77410 bronze badges
77410 bronze badges
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
9 hours ago
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 9 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
3,6182 gold badges9 silver badges25 bronze badges
add a comment |
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%2f3289009%2fwhy-does-this-quadratic-expression-have-three-zeroes%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
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
9 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
9 hours ago
3
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
9 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
9 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
9 hours ago