Find the equation of the parabola that passes through (4,-7) and has vertex (1,-6)
Hello Guest!
\(P_{v}(1,-6)\\ P_1 (4,-7) \)
Because of the symmetry
\(P_2(-2,-7)\)
\(\large y=ax^2+bx+c \)
\(c=-6\ (vertex)\)
\(-6=\ a\ +\ b\ -\ 6\) | \(\times 4\)
\(-24=4a+4b-24\) | subtr.
\(\underline{-7=16a+4b-6}\)
\(17=12a+18\\ \ a\ =\frac{17-18}{12}\)
\(a=-\frac{1}{12}\)
\(-6=-\frac{1}{12}+b-6\)
\(b=\frac{1}{12}\)
\(The\ equation\ of\ the\ parabola\ is\) \(y=-\frac{1}{12}x^2+\frac{1}{12}x-6\)
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