How do you write the quadratic function in standard form given points (-3,-4), (-1,0), (9, -10)?

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Answer 1 · 2017-03-24T09:53:28

$4 y = - x^{2} + 4 x + 5$ $4y = -x^2 +4 x +5$

$a x^{2} + b x + c = y$ $ax^2 + bx + c = y$

$3^{2} a - 3 b + c = - 4 \Rightarrow c (a, b) = 3 b - 9 a - 4$ $3^2 a - 3b + c = -4 Rightarrow c(a, b) = 3b - 9a - 4$

$a - b + c = 0$ $a - b + c = 0$

$9^{2} a + 9 b + c = - 10$ $9^2 a + 9b + c = -10$

From the second, $a - b + 3 b - 9 a - 4 = 0$ $a - b + 3b - 9a - 4 = 0$

$2 b - 8 a = 4 \Rightarrow b (a) = 4 a + 2$ $2b - 8a = 4 Rightarrow b(a) = 4a + 2$

$c = 3 (4 a + 2) - 9 a - 4$ $c = 3(4a + 2) - 9a - 4$

$c (a) = 3 a + 2$ $c(a) = 3a +2$

From the third, $81 a + 9 (4 a + 2) + 3 a + 2 = - 10$ $81a + 9(4a + 2) + 3a + 2 = -10$

$(81 + 36 + 3) a = - 10 - 18 - 2$ $(81 + 36 + 3)a = -10 - 18 - 2$

$a = - \frac{1}{4}, b (a) = 2 - 1 = 1, c (a) = 2 - \frac{3}{4} = \frac{5}{4}$ $a = - 1/4, b(a) = 2 -1 = 1, c(a) = 2-3/4 = 5/4$

$y = - \frac{1}{4} x^{2} + x + \frac{5}{4}$ $y = -1/4 x^2 +x +5/4$