Use a system of equations to find the parabola of the form yequals=axsquared2plus+bxplus+c that goes through the three given poi
1 answer:
Given that a parabola of the form 
passing through the points (2, -11), (-2, -23) and (4, -53)
Thus, substituting the points we have:
We solve equations (1), (2) and (3) simulataneously. (There are many mays it can be solved but I will use row reduction method here).
We form the augumented matrix for equations (1), (2) and (3) and perform elementary row operations as follows:
![\left[\begin{array}{ccc}4&2&1\\4&-2&1\\16&4&1\end{array}\right| \left.\begin{array}{c}-11\\-23\\-53\end{array}\right] \ \ \ \ \ \frac{1}{4} R_1\rightarrow R_1 \\ \\ \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\4&-2&1\\16&4&1\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\-23\\-53\end{array}\right] \ \ \ \ \ {{-4R_1+R_2\rightarrow R_2} \atop {-16R_1+R_3\rightarrow R_3}}](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%261%5C%5C4%26-2%261%5C%5C16%264%261%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-11%5C%5C-23%5C%5C-53%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5Cfrac%7B1%7D%7B4%7D%20R_1%5Crightarrow%20R_1%20%5C%5C%20%20%5C%5C%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%20%5Cfrac%7B1%7D%7B4%7D%20%5C%5C4%26-2%261%5C%5C16%264%261%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-%20%5Cfrac%7B11%7D%7B4%7D%20%5C%5C-23%5C%5C-53%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%7B%7B-4R_1%2BR_2%5Crightarrow%20R_2%7D%20%5Catop%20%7B-16R_1%2BR_3%5Crightarrow%20R_3%7D%7D)
![\left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&-4&0\\0&-4&-3\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\-12\\-9\end{array}\right] \ \ \ \ \ - \frac{1}{4} R_2 \\ \\ \left[\begin{array}{ccc}1& \frac{1}{2} & \frac{1}{4} \\0&1&0\\0&-4&-3\end{array}\right| \left.\begin{array}{c}- \frac{11}{4} \\3\\-9\end{array}\right] \ \ \ \ \ {{ -\frac{1}{2} R_2+R_1\rightarrow R_1} \atop {4R_2+R_3\rightarrow R_3}}](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%20%5Cfrac%7B1%7D%7B4%7D%20%5C%5C0%26-4%260%5C%5C0%26-4%26-3%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-%20%5Cfrac%7B11%7D%7B4%7D%20%5C%5C-12%5C%5C-9%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20-%20%5Cfrac%7B1%7D%7B4%7D%20R_2%20%20%5C%5C%20%20%5C%5C%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%20%5Cfrac%7B1%7D%7B4%7D%20%5C%5C0%261%260%5C%5C0%26-4%26-3%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-%20%5Cfrac%7B11%7D%7B4%7D%20%5C%5C3%5C%5C-9%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%20%7B%7B%20-%5Cfrac%7B1%7D%7B2%7D%20R_2%2BR_1%5Crightarrow%20R_1%7D%20%5Catop%20%7B4R_2%2BR_3%5Crightarrow%20R_3%7D%7D)
![\left[\begin{array}{ccc}1&0& \frac{1}{4} \\0&1&0\\0&0&-3\end{array}\right| \left.\begin{array}{c}- \frac{17}{4} \\3\\3\end{array}\right] \ \ \ \ \ - \frac{1}{3} R_3 \\ \\ \left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right| \left.\begin{array}{c}-4\\3\\-1\end{array}\right] \ \ \ \ \ - \frac{1}{4} R_3+R_1\rightarrow R_1](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%26%20%5Cfrac%7B1%7D%7B4%7D%20%5C%5C0%261%260%5C%5C0%260%26-3%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-%20%5Cfrac%7B17%7D%7B4%7D%20%5C%5C3%5C%5C3%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20-%20%5Cfrac%7B1%7D%7B3%7D%20R_3%20%5C%5C%20%20%5C%5C%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%7C%20%20%5Cleft.%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C3%5C%5C-1%5Cend%7Barray%7D%5Cright%5D%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%20-%20%5Cfrac%7B1%7D%7B4%7D%20R_3%2BR_1%5Crightarrow%20R_1)
Thus, a = -4, b = 3, c = -1
Therefore, the required polynomial is
Thus, substituting the points we have:
We solve equations (1), (2) and (3) simulataneously. (There are many mays it can be solved but I will use row reduction method here).
We form the augumented matrix for equations (1), (2) and (3) and perform elementary row operations as follows:
Thus, a = -4, b = 3, c = -1
Therefore, the required polynomial is
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Answer:
X = 33,669.78 ft
Y = 15285.76 ft
Step-by-step explanation:
Given data:
let x be the distance btwn arch and plane
and y be the distance btwn ground point and arch y
we know that
xcos27 = 30000
x = 33,669.78 ft
b) we knwo that
y = tan(27) (30000)
y = 15285.76 ft
