How do you write a quadratic equation with x-intercepts: 0,2 ; point: (3,-6) ?

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Answer 1 · 2017-05-01T08:34:04

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

The quadratic must have factors $x$ $x$ and $(x - 2)$ $(x-2)$ in order to have the required $x$ $x$ -intercepts.

So it can be written in the form:

$y = k x (x - 2)$ $y = kx(x-2)$

for some constant $k$ $k$ to be determined.

Substitute the given point $(3, - 6)$ $(3, -6)$ to find:

$- 6 = k (3) (3 - 2) = 3 k$ $color(blue)(-6) = k(color(blue)(3))(color(blue)(3)-2) = 3k$

Divide both ends by $3$ $3$ to find:

$k = - 2$ $k = -2$

So:

$y = - 2 x (x - 2) = - 2 x^{2} + 4 x$ $y = -2x(x-2) = -2x^2+4x$

graph{(4x^2+y^2-0.01)(4(x-2)^2+y^2-0.01)(4(x-3)^2+(y+6)^2-0.01)(y+2x^2-4x) = 0 [-3.6, 6.4, -7.08, 2.92]}