How do I find the formula for a parabola from two points and $y$-coordinate for the minimum?
I am trying to find the function for a parabola. The things that are known are the points $(x_1,y_1)$ $(x_3,y_3)$ and in between somewhere the parabola has its vertex, which is its minimum. The $x$-coordinate of this point is not known only the $y_2$ is known.
Now how am I supposed to find the function for the parabola?
(I know that it is $ax^2+bx+c$ but how do I calculate the $a$,$b$ and $c$ from only the 2 and a half points?)
functions
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
add a comment |
I am trying to find the function for a parabola. The things that are known are the points $(x_1,y_1)$ $(x_3,y_3)$ and in between somewhere the parabola has its vertex, which is its minimum. The $x$-coordinate of this point is not known only the $y_2$ is known.
Now how am I supposed to find the function for the parabola?
(I know that it is $ax^2+bx+c$ but how do I calculate the $a$,$b$ and $c$ from only the 2 and a half points?)
functions
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
add a comment |
I am trying to find the function for a parabola. The things that are known are the points $(x_1,y_1)$ $(x_3,y_3)$ and in between somewhere the parabola has its vertex, which is its minimum. The $x$-coordinate of this point is not known only the $y_2$ is known.
Now how am I supposed to find the function for the parabola?
(I know that it is $ax^2+bx+c$ but how do I calculate the $a$,$b$ and $c$ from only the 2 and a half points?)
functions
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
I am trying to find the function for a parabola. The things that are known are the points $(x_1,y_1)$ $(x_3,y_3)$ and in between somewhere the parabola has its vertex, which is its minimum. The $x$-coordinate of this point is not known only the $y_2$ is known.
Now how am I supposed to find the function for the parabola?
(I know that it is $ax^2+bx+c$ but how do I calculate the $a$,$b$ and $c$ from only the 2 and a half points?)
functions
functions
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
edited Nov 27 '18 at 20:10
Hans Hüttel
3,1972921
3,1972921
asked Nov 27 '18 at 20:06
wizzi
1
asked Nov 27 '18 at 20:06
wizzi
1
asked Nov 27 '18 at 20:06
wizzi
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You have two equations:
$$y_1 = a{x_1}^2 + bx_1 + c$$
$$y_3 = a{x_3}^2 + bx_3 + c$$
and also know that the $x$-coordinate of the vertex is $x_2 = -frac{b}{2a}$, so
$$y_2 = a{x_2}^2 + bx_2 + c$$
Substitute $-frac{b}{2a}$ for $x_2$ in the above. You now have three equations whose unknowns are $a$, $b$ and $c$.
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
add a comment |
Another approach: Let $y=a(x-x_2)^2+y_2iff y-y_2=a(x-x_2)^2$. Plugging in the coordinates and dividing both equations gives
$$frac{y_3-y_2}{y_1-y_2}=frac{(x_3-x_2)^2}{(x_1-x_2)^2}.$$
Now convince yourself that the left hand side is positive, hence $x_2$ is easily calculated from
$$sqrt{frac{y_3-y_2}{y_1-y_2}}=pmfrac{(x_3-x_2)}{(x_1-x_2)}.$$
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
add a comment |
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2 Answers
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2 Answers
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You have two equations:
$$y_1 = a{x_1}^2 + bx_1 + c$$
$$y_3 = a{x_3}^2 + bx_3 + c$$
and also know that the $x$-coordinate of the vertex is $x_2 = -frac{b}{2a}$, so
$$y_2 = a{x_2}^2 + bx_2 + c$$
Substitute $-frac{b}{2a}$ for $x_2$ in the above. You now have three equations whose unknowns are $a$, $b$ and $c$.
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
add a comment |
You have two equations:
$$y_1 = a{x_1}^2 + bx_1 + c$$
$$y_3 = a{x_3}^2 + bx_3 + c$$
and also know that the $x$-coordinate of the vertex is $x_2 = -frac{b}{2a}$, so
$$y_2 = a{x_2}^2 + bx_2 + c$$
Substitute $-frac{b}{2a}$ for $x_2$ in the above. You now have three equations whose unknowns are $a$, $b$ and $c$.
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
add a comment |
You have two equations:
$$y_1 = a{x_1}^2 + bx_1 + c$$
$$y_3 = a{x_3}^2 + bx_3 + c$$
and also know that the $x$-coordinate of the vertex is $x_2 = -frac{b}{2a}$, so
$$y_2 = a{x_2}^2 + bx_2 + c$$
Substitute $-frac{b}{2a}$ for $x_2$ in the above. You now have three equations whose unknowns are $a$, $b$ and $c$.
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
You have two equations:
$$y_1 = a{x_1}^2 + bx_1 + c$$
$$y_3 = a{x_3}^2 + bx_3 + c$$
and also know that the $x$-coordinate of the vertex is $x_2 = -frac{b}{2a}$, so
$$y_2 = a{x_2}^2 + bx_2 + c$$
Substitute $-frac{b}{2a}$ for $x_2$ in the above. You now have three equations whose unknowns are $a$, $b$ and $c$.
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
answered Nov 27 '18 at 20:16
Hans Hüttel
3,1972921
3,1972921
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
add a comment |
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
Thanks for the answers. But could you explain how you know that what x_2 equals?
– wizzi
Nov 30 '18 at 15:25
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
See e.g. mathwarehouse.com/geometry/parabola/vertex-of-a-parabola.php.
– Hans Hüttel
Nov 30 '18 at 17:34
add a comment |
Another approach: Let $y=a(x-x_2)^2+y_2iff y-y_2=a(x-x_2)^2$. Plugging in the coordinates and dividing both equations gives
$$frac{y_3-y_2}{y_1-y_2}=frac{(x_3-x_2)^2}{(x_1-x_2)^2}.$$
Now convince yourself that the left hand side is positive, hence $x_2$ is easily calculated from
$$sqrt{frac{y_3-y_2}{y_1-y_2}}=pmfrac{(x_3-x_2)}{(x_1-x_2)}.$$
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
add a comment |
Another approach: Let $y=a(x-x_2)^2+y_2iff y-y_2=a(x-x_2)^2$. Plugging in the coordinates and dividing both equations gives
$$frac{y_3-y_2}{y_1-y_2}=frac{(x_3-x_2)^2}{(x_1-x_2)^2}.$$
Now convince yourself that the left hand side is positive, hence $x_2$ is easily calculated from
$$sqrt{frac{y_3-y_2}{y_1-y_2}}=pmfrac{(x_3-x_2)}{(x_1-x_2)}.$$
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
add a comment |
Another approach: Let $y=a(x-x_2)^2+y_2iff y-y_2=a(x-x_2)^2$. Plugging in the coordinates and dividing both equations gives
$$frac{y_3-y_2}{y_1-y_2}=frac{(x_3-x_2)^2}{(x_1-x_2)^2}.$$
Now convince yourself that the left hand side is positive, hence $x_2$ is easily calculated from
$$sqrt{frac{y_3-y_2}{y_1-y_2}}=pmfrac{(x_3-x_2)}{(x_1-x_2)}.$$
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
Another approach: Let $y=a(x-x_2)^2+y_2iff y-y_2=a(x-x_2)^2$. Plugging in the coordinates and dividing both equations gives
$$frac{y_3-y_2}{y_1-y_2}=frac{(x_3-x_2)^2}{(x_1-x_2)^2}.$$
Now convince yourself that the left hand side is positive, hence $x_2$ is easily calculated from
$$sqrt{frac{y_3-y_2}{y_1-y_2}}=pmfrac{(x_3-x_2)}{(x_1-x_2)}.$$
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
answered Nov 27 '18 at 20:44
Michael Hoppe
10.8k31834
10.8k31834
add a comment |
add a comment |
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