A question on the parabola
$f:[-2,5] to mathbb{R}$
$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.
algebra-precalculus
edited Dec 9 at 15:45
mm-crj
382212
asked Dec 9 at 15:25
Hatm00
276
add a comment |
$f:[-2,5] to mathbb{R}$
$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.
algebra-precalculus
edited Dec 9 at 15:45
mm-crj
382212
asked Dec 9 at 15:25
Hatm00
276
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45
add a comment |
$f:[-2,5] to mathbb{R}$
$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.
algebra-precalculus
edited Dec 9 at 15:45
mm-crj
382212
asked Dec 9 at 15:25
Hatm00
276
$f:[-2,5] to mathbb{R}$
$f(x)= x^2-2mx+m+2$ The smallest value that the function can take is $-10$. Find the sum of the values that $m$ can take.
algebra-precalculus
algebra-precalculus
edited Dec 9 at 15:45
mm-crj
382212
asked Dec 9 at 15:25
Hatm00
276
edited Dec 9 at 15:45
mm-crj
382212
asked Dec 9 at 15:25
Hatm00
276
edited Dec 9 at 15:45
mm-crj
382212
edited Dec 9 at 15:45
mm-crj
382212
edited Dec 9 at 15:45
mm-crj
382212
382212
asked Dec 9 at 15:25
Hatm00
276
asked Dec 9 at 15:25
Hatm00
276
asked Dec 9 at 15:25
Hatm00
276
276
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45
add a comment |
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45
add a comment |
2 Answers
2
active
oldest
votes
The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)
However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.
Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$
- If $-2<m<5$, the minimum value is taken at $m$
- if $mle-2$, the minimum is taken at …
- if $mge5$, the minimum is taken at …
We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.
We have $f(-2)=-10$ if and only if $m=-16/5<-2$.
We have $f(5)=-10$ if and only if $m=37/9$.
Two of these values have to be discarded.
answered Dec 9 at 15:28
egreg
178k1484201
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
|
show 1 more comment
The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Since you have $a = 1 > 0$, then a minimum exists.
$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$
$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$
$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$
This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.
answered Dec 9 at 15:35
KM101
4,836421
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
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%2f3032495%2fa-question-on-the-parabola%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
The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)
However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.
Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$
- If $-2<m<5$, the minimum value is taken at $m$
- if $mle-2$, the minimum is taken at …
- if $mge5$, the minimum is taken at …
We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.
We have $f(-2)=-10$ if and only if $m=-16/5<-2$.
We have $f(5)=-10$ if and only if $m=37/9$.
Two of these values have to be discarded.
answered Dec 9 at 15:28
egreg
178k1484201
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
|
show 1 more comment
The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)
However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.
Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$
- If $-2<m<5$, the minimum value is taken at $m$
- if $mle-2$, the minimum is taken at …
- if $mge5$, the minimum is taken at …
We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.
We have $f(-2)=-10$ if and only if $m=-16/5<-2$.
We have $f(5)=-10$ if and only if $m=37/9$.
Two of these values have to be discarded.
answered Dec 9 at 15:28
egreg
178k1484201
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
|
show 1 more comment
The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)
However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.
Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$
- If $-2<m<5$, the minimum value is taken at $m$
- if $mle-2$, the minimum is taken at …
- if $mge5$, the minimum is taken at …
We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.
We have $f(-2)=-10$ if and only if $m=-16/5<-2$.
We have $f(5)=-10$ if and only if $m=37/9$.
Two of these values have to be discarded.
answered Dec 9 at 15:28
egreg
178k1484201
The smallest value of $ax^2+bx+c$ (when $a>0$) is taken at $x$ halfway between the roots, which is
$$
x=-frac{b}{2a}=m
$$
(For $a<0$, that abscissa gives the maximum value.)
However, your problem is a bit more complicated, because you bound the function over $[-2,5]$. Then the minimum value can be at $m$ (if it is in this range), at $-2$ or at $5$.
Now you have $f(-2)=5m+6$ and $f(5)=-9m+27$. Also
$$
f(m)=m^2-2m^2+m+2=-m^2+m+2
$$
- If $-2<m<5$, the minimum value is taken at $m$
- if $mle-2$, the minimum is taken at …
- if $mge5$, the minimum is taken at …
We have $f(m)=-10$ if and only if $m^2-m-12=0$, that is $m=4$ or $m=-3$.
We have $f(-2)=-10$ if and only if $m=-16/5<-2$.
We have $f(5)=-10$ if and only if $m=37/9$.
Two of these values have to be discarded.
answered Dec 9 at 15:28
egreg
178k1484201
edited Dec 9 at 16:19
answered Dec 9 at 15:28
egreg
178k1484201
answered Dec 9 at 15:28
egreg
178k1484201
answered Dec 9 at 15:28
egreg
178k1484201
178k1484201
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
|
show 1 more comment
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
When the root exists, if the first claim aims to be general (since we are in a precalculus context). Or are you assuming that we can refer also to the compelx roots in that context?
– gimusi
Dec 9 at 15:32
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
Hey smallest value can also occur at boundary right ?
– rsadhvika
Dec 9 at 15:34
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
But it's a delimited function. It can be equal f(-2) or f(5). Is my thought wrong?
– Hatm00
Dec 9 at 15:42
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@gimusi Shifting the parabola up or down doesn't change the point where the minimum is taken, only the minimum value. The halfway point doesn't change with such a translation.
– egreg
Dec 9 at 16:04
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
@Hatm00 I added further hints
– egreg
Dec 9 at 16:19
|
show 1 more comment
The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Since you have $a = 1 > 0$, then a minimum exists.
$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$
$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$
$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$
This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.
answered Dec 9 at 15:35
KM101
4,836421
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
add a comment |
The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Since you have $a = 1 > 0$, then a minimum exists.
$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$
$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$
$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$
This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.
answered Dec 9 at 15:35
KM101
4,836421
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
add a comment |
The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Since you have $a = 1 > 0$, then a minimum exists.
$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$
$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$
$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$
This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.
answered Dec 9 at 15:35
KM101
4,836421
The maximum or minimum of a parabola occurs at its vertex at $(h, k)$.
$$h = frac{-b}{2a}$$
$$k = c-frac{b^2}{4a}$$
Since you have $a = 1 > 0$, then a minimum exists.
$$k = -10 iff c-frac{b^2}{4a} = -10 iff (m+2)-frac{4m^2}{4} = -10$$
$$frac{4m+8-4m^2}{4} = -10 iff -4m^2+4m+8 = -40$$
$$-4m^2+4m+48 = 0 iff m^2-m-12 = 0 implies (m-4)(m+3) = 0$$
This yields $m = 4$ and $m = -3$, but only the former is applicable given the limited domain.
answered Dec 9 at 15:35
KM101
4,836421
edited Dec 9 at 15:47
answered Dec 9 at 15:35
KM101
4,836421
answered Dec 9 at 15:35
KM101
4,836421
answered Dec 9 at 15:35
KM101
4,836421
4,836421
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
add a comment |
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
But we don't know is vertex in definition range.
– Hatm00
Dec 9 at 15:39
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
@Hatm00 I’ve edited the answer. (I misread the question.)
– KM101
Dec 9 at 15:51
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
Thanks so much for your help. If we add root with the boundary values we can reach the result.
– Hatm00
Dec 9 at 16:01
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
It would be good if the downvoter clarifies if the answer is missing anything.
– KM101
Dec 9 at 16:03
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3032495%2fa-question-on-the-parabola%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
The smallest value occurs at the vertex of the parabola. Do you know how to find the vertex?
– rogerl
Dec 9 at 15:27
You do know you can write text outside the dollar signs, right? You don't have to have a backslash before every space, and it looks better.
– Arthur
Dec 9 at 15:28
@rogerl I tried but it didn't work out.
– Hatm00
Dec 9 at 15:33
@Hatm00 keep in mind the minimum value can also occur at boundary, see
– rsadhvika
Dec 9 at 15:42
You're right but also I found the values that you found and vertex value. As I said it didn't give the result.
– Hatm00
Dec 9 at 15:45