Finding coefficients of a quadratic equation looking to this plot.

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For an equation $(ax^2+bx+c)e^-x$ I must find the three coefficients $a$, $b$ and $c$ by looking at this plot

I have the feeling I need to find three equations in order to form a coherent system but I only find these two:

$$0=0^2a+0b+c$$
$$16=4a+2b+c$$

I easily find out that $b=8-2a$ and $c=0$, but then what? I've been working on it much more than it surely deserves.

systems-of-equations graphing-functions quadratics problem-solving

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edited Sep 9 at 17:22

Robert Z

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asked Sep 9 at 17:09

torito verdejo

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up vote
3
down vote

favorite

1

For an equation $(ax^2+bx+c)e^-x$ I must find the three coefficients $a$, $b$ and $c$ by looking at this plot

I have the feeling I need to find three equations in order to form a coherent system but I only find these two:

$$0=0^2a+0b+c$$
$$16=4a+2b+c$$

I easily find out that $b=8-2a$ and $c=0$, but then what? I've been working on it much more than it surely deserves.

systems-of-equations graphing-functions quadratics problem-solving

share|cite|improve this question

edited Sep 9 at 17:22

Robert Z

85.5k1055123

asked Sep 9 at 17:09

torito verdejo

635

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

For an equation $(ax^2+bx+c)e^-x$ I must find the three coefficients $a$, $b$ and $c$ by looking at this plot

I have the feeling I need to find three equations in order to form a coherent system but I only find these two:

$$0=0^2a+0b+c$$
$$16=4a+2b+c$$

I easily find out that $b=8-2a$ and $c=0$, but then what? I've been working on it much more than it surely deserves.

systems-of-equations graphing-functions quadratics problem-solving

share|cite|improve this question

edited Sep 9 at 17:22

Robert Z

85.5k1055123

asked Sep 9 at 17:09

torito verdejo

635

For an equation $(ax^2+bx+c)e^-x$ I must find the three coefficients $a$, $b$ and $c$ by looking at this plot

I have the feeling I need to find three equations in order to form a coherent system but I only find these two:

$$0=0^2a+0b+c$$
$$16=4a+2b+c$$

I easily find out that $b=8-2a$ and $c=0$, but then what? I've been working on it much more than it surely deserves.

systems-of-equations graphing-functions quadratics problem-solving

systems-of-equations graphing-functions quadratics problem-solving

share|cite|improve this question

edited Sep 9 at 17:22

Robert Z

85.5k1055123

asked Sep 9 at 17:09

torito verdejo

635

share|cite|improve this question

edited Sep 9 at 17:22

Robert Z

85.5k1055123

asked Sep 9 at 17:09

torito verdejo

635

share|cite|improve this question

share|cite|improve this question

edited Sep 9 at 17:22

Robert Z

85.5k1055123

edited Sep 9 at 17:22

Robert Z

85.5k1055123

edited Sep 9 at 17:22

Robert Z

85.5k1055123

85.5k1055123

asked Sep 9 at 17:09

torito verdejo

635

asked Sep 9 at 17:09

torito verdejo

635

asked Sep 9 at 17:09

torito verdejo

635

635

add a comment | 

add a comment | 

2 Answers
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up vote
7
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accepted

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^-x$, then
$$f'(x)=(2ax+b-ax^2-bx-c)e^-x$$
and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

share|cite|improve this answer

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

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up vote
5
down vote

You also have $f(x) geq 0$ for any $x in mathbbR$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^-x$$

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answered Sep 9 at 17:14

LucaMac

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote



accepted

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^-x$, then
$$f'(x)=(2ax+b-ax^2-bx-c)e^-x$$
and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

share|cite|improve this answer

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

add a comment | 

up vote
7
down vote



accepted

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^-x$, then
$$f'(x)=(2ax+b-ax^2-bx-c)e^-x$$
and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

share|cite|improve this answer

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^-x$, then
$$f'(x)=(2ax+b-ax^2-bx-c)e^-x$$
and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

share|cite|improve this answer

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

You are not considering the fact that $0$ is a minimum point and $2$ is a maximum point. Let $f(x)=(ax^2+bx+c)e^-x$, then
$$f'(x)=(2ax+b-ax^2-bx-c)e^-x$$
and therefore $f'(0)=f'(2)=0$. Note that $f'(0)=b-c=0$ implies that $b=c=0$. Finally from $b=8-2a$ it follows that $a=4$.

share|cite|improve this answer

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

share|cite|improve this answer

share|cite|improve this answer

edited Sep 9 at 18:09

edited Sep 9 at 18:09

edited Sep 9 at 18:09

answered Sep 9 at 17:15

Robert Z

85.5k1055123

answered Sep 9 at 17:15

Robert Z

85.5k1055123

answered Sep 9 at 17:15

Robert Z

85.5k1055123

85.5k1055123

add a comment | 

add a comment | 

up vote
5
down vote

You also have $f(x) geq 0$ for any $x in mathbbR$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^-x$$

share|cite|improve this answer

answered Sep 9 at 17:14

LucaMac

1,25414

add a comment | 

up vote
5
down vote

You also have $f(x) geq 0$ for any $x in mathbbR$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^-x$$

share|cite|improve this answer

answered Sep 9 at 17:14

LucaMac

1,25414

add a comment | 

up vote
5
down vote

up vote
5
down vote

You also have $f(x) geq 0$ for any $x in mathbbR$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^-x$$

share|cite|improve this answer

answered Sep 9 at 17:14

LucaMac

1,25414

You also have $f(x) geq 0$ for any $x in mathbbR$, and since $c=0$ this is equivalent to $b=0$. So ($a=4$) we have $$f(x) = 4x^2e^-x$$

share|cite|improve this answer

answered Sep 9 at 17:14

LucaMac

1,25414

share|cite|improve this answer

share|cite|improve this answer

answered Sep 9 at 17:14

LucaMac

1,25414

answered Sep 9 at 17:14

LucaMac

1,25414

answered Sep 9 at 17:14

LucaMac

1,25414

1,25414

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