The equation of the parabola whose axis is ve | vedclass

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Question

(A) The general equation of a parabola with a vertical axis is $y = ax^2 + bx + c$.Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 + b(

Vedclass · Accepted Answer

$x^2 - 3x - y = 0$

Vedclass · Answer

(A) The general equation of a parabola with a vertical axis is $y = ax^2 + bx + c$.Since the parabola passes through $(0, 0)$,we have $0 = a(0)^2 + b(0) + c$,which gives $c = 0$.Since it passes through $(3, 0)$,we have $0 = a(3)^2 + b(3) + 0$,which simplifies to $9a + 3b = 0$,or $3a + b = 0$,so $b = -3a$.Since it passes through $(-1, 4)$,we have $4 = a(-1)^2 + b(-1) + 0$,which gives $4 = a - b$.Substituting $b = -3a$ into the equation $4 = a - b$,we get $4 = a - (-3a) = 4a$,which implies $a = 1$.Then $b = -3(1) = -3$.Thus,the equation is $y = x^2 - 3x$,which can be rewritten as $x^2 - 3x - y = 0$.

The equation of the parabola whose axis is vertical and passes through the points $(0, 0), (3, 0)$ and $(-1, 4)$ is

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