Suppose a parabola passes through (0,4), (1,9 | vedclass

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(A) Let the equation of the parabola be $y = ax^2 + bx + c$.Given that it passes through $(0,4), (1,9),$ and $(4,5)$.Substituting $(0,4)$ into the equ

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$19x^2 + 12y - 79x - 48 = 0$

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(A) Let the equation of the parabola be $y = ax^2 + bx + c$.Given that it passes through $(0,4), (1,9),$ and $(4,5)$.Substituting $(0,4)$ into the equation: $4 = a(0)^2 + b(0) + c \implies c = 4$.Substituting $(1,9)$ into the equation: $9 = a(1)^2 + b(1) + 4 \implies a + b = 5 \dots (1)$.Substituting $(4,5)$ into the equation: $5 = a(4)^2 + b(4) + 4 \implies 16a + 4b = 1 \dots (2)$.Multiplying equation $(1)$ by $4$: $4a + 4b = 20 \dots (3)$.Subtracting equation $(3)$ from equation $(2)$: $(16a - 4a) = 1 - 20 \implies 12a = -19 \implies a = -\frac{19}{12}$.Substituting $a$ into equation $(1)$: $-\frac{19}{12} + b = 5 \implies b = 5 + \frac{19}{12} = \frac{60+19}{12} = \frac{79}{12}$.Substituting $a, b, c$ into the general equation: $y = -\frac{19}{12}x^2 + \frac{79}{12}x + 4$.Multiplying by $12$: $12y = -19x^2 + 79x + 48$.Rearranging terms: $19x^2 + 12y - 79x - 48 = 0$.

Suppose a parabola passes through $(0,4), (1,9)$ and $(4,5)$ and has its axis parallel to the $y$-axis. Then the equation of the parabola is

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