Mathematics, 24.07.2019 06:20 maybe67
Use a system of equations to find the parabola of the form yequals=axsquared2plus+bxplus+c that goes through the three given points. (22, negative 11−11), (negative 2−2, negative 23−23), (44, negative 53−53)
Answers: 1
passing through the points (2, -11), (-2, -23) and (4, -53)
![\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}}](/tpl/images/0126/2703/d18fb.png)
![\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}}](/tpl/images/0126/2703/14da7.png)
![\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](/tpl/images/0126/2703/4b373.png)
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Use a system of equations to find the parabola of the form yequals=axsquared2plus+bxplus+c that goes...
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