Question

Find quadratic equation from 3 points calculator

Answer 1

To find the quadratic equation from 3 points, use the method of substitution and solve the system of equations. Substituting the given points into the general quadratic equation will result in a system of equations that can be solved for the quadratic equation.

Step-by-step explanation:

To find the quadratic equation from 3 points, we can use the method of substitution. Let's say the three points are (x1, y1), (x2, y2), and (x3, y3). We can substitute these points into the general quadratic equation, y = ax^2 + bx + c, to get three equations. Solving this system of equations will give us the values of a, b, and c, and thus the quadratic equation.

For example, let's say the three points are (1, 3), (2, 6), and (3, 11). Substituting these points into the general quadratic equation, we get the following equations:

3 = a(1^2) + b(1) + c

6 = a(2^2) + b(2) + c

11 = a(3^2) + b(3) + c

Solving this system of equations will give us the values of a, b, and c, and thus the quadratic equation.

Answer 2

Given the points P(-2, 7), Q(6, 8) and R(-1, -3), we find the quadratic equation of the form
`y=ax^2+bx+c` as follows:

Step 1:
We check to make sure that no pair of points is colinear (i.e. lie on the same line) by making that the slopes of the combinations of the line are not equal.

Slope of PQ =
`(8-7)/(6-(-2)) = (1)/(8)`
Slope of PR =
`(-3-7)/(-1-(-2)) = (-10)/(1) =-10`
Slope of QR =
`(-3-8)/(-1-6) = (-11)/(-7) = (11)/(7)`

Since, the slope of the lines are not equal, thus the points are not colinear nor parallel.

Step 2:
Substitute the x-values and the y-values of the given points into the quadratic equation formular to have three system of equations as follows:


`a(-2)^2+b(-2)+c=7\Rightarrow 4a-2b+c=7 \ .\ .\ .\ (1) \\ \\ a(6)^2+b(6)+c=8\Rightarrow36a+6b+c=8\ .\ .\ .\ (2) \\ \\ a(-1)^2+b(-1)+c=-3\Rightarrow a-b+c=-3\ .\ .\ .\ (3)`

Step 3:
We solve the system of equations as follows:

`\left[\begin{array}{ccccc}4&-2&1&|&7\\36&6&1&|&8\\1&-1&1&|&-3\end{array}\right] \ \ \ R_1\leftrightarrow R_3 \\ \\ \left[\begin{array}{ccccc}1&-1&1&|&-3\\36&6&1&|&8\\4&-2&1&|&7\end{array}\right] \ \ \ {{-36R_1+R_2\rightarrow R_2} \atop {-4R_1+R_3\rightarrow R_3}} \\ \\ \left[\begin{array}{ccccc}1&-1&1&|&-3\\0&42&-35&|&116\\0&2&-3&|&19\end{array}\right] \ \ \ R_2\leftrightarrow R_3`


`\left[\begin{array}{ccccc}1&-1&1&|&-3\\0&2&-3&|&19\\0&42&-35&|&116\end{array}\right] \ \ \ (1)/(2) R_2\leftrightarrow R_2 \\ \\ \left[\begin{array}{ccccc}1&-1&1&|&-3\\0&1&(-3)/(2)&|&(19)/(2)\\0&42&-35&|&116\end{array}\right] \ \ \ {{R_1+R_2\rightarrow R_1} \atop {-42R_2+R_3\rightarrow R_3}} \\ \\ \left[\begin{array}{ccccc}1&0&(-1)/(2)&|&(13)/(2)\\0&1&(-3)/(2)&|&(19)/(2)\\0&0&28&|&-283\end{array}\right] \ \ \ (1)/(28) R_3\leftrightarrow R_3`


`\left[\begin{array}{ccccc}1&0&(-1)/(2)&|&(13)/(2)\\0&1&(-3)/(2)&|&(19)/(2)\\0&0&1&|&(-283)/(28)\end{array}\right] \ \ \ {{(1)/(2)R_3+R_1\rightarrow R_1} \atop {(3)/(2)R_3+R_2\rightarrow R_2}} \\ \\ \left[\begin{array}{ccccc}1&0&0&|&(81)/(56)\\0&1&0&|&(-317)/(56)\\0&0&1&|&(-283)/(28)\end{array}\right]`

Thus,


`a=(81)/(56); \ \ \ b=(-317)/(56); \ \ \ c=(-283)/(28)`

Therefore, the quadratic equation with the three given points is


`y=(81)/(56)x^2-(317)/(56)x-(283)/(28)`