Find the equation of the parabola whose axis | vedclass

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

Vedclass · Accepted Answer

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

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(C) The general equation of a parabola with its axis parallel to the $y$-axis is given by $y = Ax^2 + Bx + C$ ...$(i)$Since the parabola passes through the points $(0, 4), (1, 9),$ and $(4, 5)$,these points must satisfy the equation.Substituting $(0, 4)$ in $(i)$: $4 = A(0)^2 + B(0) + C \Rightarrow C = 4$ ...(ii)Substituting $(1, 9)$ in $(i)$: $9 = A(1)^2 + B(1) + 4 \Rightarrow A + B = 5$ ...(iii)Substituting $(4, 5)$ in $(i)$: $5 = A(4)^2 + B(4) + 4 \Rightarrow 16A + 4B = 1$ ...(iv)Dividing (iv) by $4$,we get $4A + B = 0.25$ or $4A + B = \frac{1}{4}$ ...$(v)$Subtracting (iii) from $(v)$: $(4A + B) - (A + B) = \frac{1}{4} - 5$ $\Rightarrow 3A = -\frac{19}{4}$ $\Rightarrow A = -\frac{19}{12}$Substituting $A$ in (iii): $-\frac{19}{12} + B = 5 \Rightarrow B = 5 + \frac{19}{12} = \frac{79}{12}$Thus,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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