Find the equation of the parabola whose axis | vedclass

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Question

(C) The general equation of a parabola with an axis parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$ ...$(i)$Since the parabola passes through

Vedclass · Accepted Answer

$y=\frac{-19}{12} x^2+\frac{79}{12} x+4$

Vedclass · Answer

(C) The general equation of a parabola with an axis parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$ ...$(i)$Since the parabola passes through $(0,4)$,we have $4 = A(0)^2 + B(0) + C$,which gives $C = 4$ ...(ii)Since it passes through $(1,9)$,we have $9 = A(1)^2 + B(1) + 4$,which simplifies to $A + B = 5$ ...(iii)Since it passes through $(4,5)$,we have $5 = A(4)^2 + B(4) + 4$,which simplifies to $16A + 4B = 1$ or $4A + B = \frac{1}{4}$ ...(iv)Subtracting Eq. (iii) from Eq. (iv): $(4A + B) - (A + B) = \frac{1}{4} - 5$ $\Rightarrow 3A = \frac{-19}{4}$ $\Rightarrow A = \frac{-19}{12}$Substituting $A$ into Eq. (iii): $\frac{-19}{12} + B = 5 \Rightarrow B = 5 + \frac{19}{12} = \frac{79}{12}$Substituting $A, B, C$ into Eq. $(i)$,the equation is $y = \frac{-19}{12} x^2 + \frac{79}{12} x + 4$.

Find the equation of the parabola whose axis is parallel to the $y$-axis and which passes through the points $(0,4), (1,9)$ and $(4,5)$.

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