Mixing
Solving an equation for 3 unknowns
$begingroup$
So I have the function $y=-a(x+b)^2+c$
And I also know that this equation goes through these three points
$(0,0), (6,12), (2,8)$
I tried to make a system with $3$ equations from this info, but I was unable to solve it.
functions systems-of-equations graphing-functions
edited Jan 30 at 17:22
the_fox
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
$endgroup$
add a comment |
$begingroup$
So I have the function $y=-a(x+b)^2+c$
And I also know that this equation goes through these three points
$(0,0), (6,12), (2,8)$
I tried to make a system with $3$ equations from this info, but I was unable to solve it.
functions systems-of-equations graphing-functions
edited Jan 30 at 17:22
the_fox
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
$endgroup$
1
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
1
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18
add a comment |
$begingroup$
So I have the function $y=-a(x+b)^2+c$
And I also know that this equation goes through these three points
$(0,0), (6,12), (2,8)$
I tried to make a system with $3$ equations from this info, but I was unable to solve it.
functions systems-of-equations graphing-functions
edited Jan 30 at 17:22
the_fox
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
$endgroup$
So I have the function $y=-a(x+b)^2+c$
And I also know that this equation goes through these three points
$(0,0), (6,12), (2,8)$
I tried to make a system with $3$ equations from this info, but I was unable to solve it.
functions systems-of-equations graphing-functions
functions systems-of-equations graphing-functions
edited Jan 30 at 17:22
the_fox
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
edited Jan 30 at 17:22
the_fox
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
edited Jan 30 at 17:22
the_fox
2,90231538
edited Jan 30 at 17:22
the_fox
2,90231538
edited Jan 30 at 17:22
the_fox
2,90231538
2,90231538
asked Jan 30 at 16:03
user655941user655941
11
asked Jan 30 at 16:03
user655941user655941
11
asked Jan 30 at 16:03
user655941user655941
11
11
1
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
1
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18
add a comment |
1
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
1
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18
1
1
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
1
1
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: The system is given by
$$0=-a(0+b)^2+c$$
$$12=-a(6+b)^2+c$$
$$8=-a(2+b)^2+c$$
From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get
$$frac{8}{b^2-(2+b)^2}=a=frac{12}{b^2-(6+b)^2}$$
Can you finish now?
Simplifying we get
$$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
add a comment |
$begingroup$
Hint:
Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that
begin{align}begin{cases}
p(6)=p(2)=0 ,\
p(0)=1,
end{cases}qquadbegin{cases}
q(0)=q(6)=0 ,\
q(2)=1,
end{cases} qquadbegin{cases}
r(2)=r(0)=0 ,\
r(6)=1,
end{cases} end{align}
and take
$$y(x)=0cdot p(x)+ 8cdot q(x)+12 cdot r(x).$$
Then $-b:$ is the abscissa of the axis if symmetry of the resulting parabola.
answered Jan 30 at 17:39
BernardBernard
123k741117
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
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oldest
votes
$begingroup$
Hint: The system is given by
$$0=-a(0+b)^2+c$$
$$12=-a(6+b)^2+c$$
$$8=-a(2+b)^2+c$$
From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get
$$frac{8}{b^2-(2+b)^2}=a=frac{12}{b^2-(6+b)^2}$$
Can you finish now?
Simplifying we get
$$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
add a comment |
$begingroup$
Hint: The system is given by
$$0=-a(0+b)^2+c$$
$$12=-a(6+b)^2+c$$
$$8=-a(2+b)^2+c$$
From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get
$$frac{8}{b^2-(2+b)^2}=a=frac{12}{b^2-(6+b)^2}$$
Can you finish now?
Simplifying we get
$$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
add a comment |
$begingroup$
Hint: The system is given by
$$0=-a(0+b)^2+c$$
$$12=-a(6+b)^2+c$$
$$8=-a(2+b)^2+c$$
From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get
$$frac{8}{b^2-(2+b)^2}=a=frac{12}{b^2-(6+b)^2}$$
Can you finish now?
Simplifying we get
$$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
$endgroup$
Hint: The system is given by
$$0=-a(0+b)^2+c$$
$$12=-a(6+b)^2+c$$
$$8=-a(2+b)^2+c$$
From the first equation we get $$c=ab^2$$ plugging this in equation (2) and (3) and solving for $$a$$ we get
$$frac{8}{b^2-(2+b)^2}=a=frac{12}{b^2-(6+b)^2}$$
Can you finish now?
Simplifying we get
$$2(12b+36)=3(4b+4)$$ This equation must be solved for $b$, it is linear in $$b$$
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
edited Jan 30 at 17:22
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
answered Jan 30 at 16:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.4k42867
78.4k42867
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
add a comment |
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
$begingroup$
Yes i tried to solve those for finding the a b and c but i was unable to do so
$endgroup$
– user655941
Jan 30 at 17:17
add a comment |
$begingroup$
Hint:
Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that
begin{align}begin{cases}
p(6)=p(2)=0 ,\
p(0)=1,
end{cases}qquadbegin{cases}
q(0)=q(6)=0 ,\
q(2)=1,
end{cases} qquadbegin{cases}
r(2)=r(0)=0 ,\
r(6)=1,
end{cases} end{align}
and take
$$y(x)=0cdot p(x)+ 8cdot q(x)+12 cdot r(x).$$
Then $-b:$ is the abscissa of the axis if symmetry of the resulting parabola.
answered Jan 30 at 17:39
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that
begin{align}begin{cases}
p(6)=p(2)=0 ,\
p(0)=1,
end{cases}qquadbegin{cases}
q(0)=q(6)=0 ,\
q(2)=1,
end{cases} qquadbegin{cases}
r(2)=r(0)=0 ,\
r(6)=1,
end{cases} end{align}
and take
$$y(x)=0cdot p(x)+ 8cdot q(x)+12 cdot r(x).$$
Then $-b:$ is the abscissa of the axis if symmetry of the resulting parabola.
answered Jan 30 at 17:39
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Hint:
Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that
begin{align}begin{cases}
p(6)=p(2)=0 ,\
p(0)=1,
end{cases}qquadbegin{cases}
q(0)=q(6)=0 ,\
q(2)=1,
end{cases} qquadbegin{cases}
r(2)=r(0)=0 ,\
r(6)=1,
end{cases} end{align}
and take
$$y(x)=0cdot p(x)+ 8cdot q(x)+12 cdot r(x).$$
Then $-b:$ is the abscissa of the axis if symmetry of the resulting parabola.
answered Jan 30 at 17:39
BernardBernard
123k741117
$endgroup$
Hint:
Find quadratic polynomials $p(x)$, $q(x)$, $r(x)$ such that
begin{align}begin{cases}
p(6)=p(2)=0 ,\
p(0)=1,
end{cases}qquadbegin{cases}
q(0)=q(6)=0 ,\
q(2)=1,
end{cases} qquadbegin{cases}
r(2)=r(0)=0 ,\
r(6)=1,
end{cases} end{align}
and take
$$y(x)=0cdot p(x)+ 8cdot q(x)+12 cdot r(x).$$
Then $-b:$ is the abscissa of the axis if symmetry of the resulting parabola.
answered Jan 30 at 17:39
BernardBernard
123k741117
answered Jan 30 at 17:39
BernardBernard
123k741117
answered Jan 30 at 17:39
BernardBernard
123k741117
answered Jan 30 at 17:39
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
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1
$begingroup$
Which equation system do you find?
$endgroup$
– Dr. Sonnhard Graubner
Jan 30 at 16:04
1
$begingroup$
Welcome to MSE. Please show us the three equations you came up with, and your attempt at solving them. Otherwise, you're just asking for someone to do your homework, which is not what this site is about.
$endgroup$
– saulspatz
Jan 30 at 16:07
$begingroup$
If you want to see my scribbles all over pages to solve it ur welcome just ask for it
$endgroup$
– user655941
Jan 30 at 17:18