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What is the equation of the parabola that has a vertex at $(- 2, - 4)$ $(-2, -4)$ and passes through point $(- 3, - 5)$ $(-3,-5)$ ?

1 Answer

Dec 7, 2015

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

Explanation:

The general vertex form of a parabola with vertex at $(a, b)$ $(a,b)$ is
$XXX y = m {(x - a)}^{2} + b XXX$ $color(white)("XXX")y=m(x-a)^2+bcolor(white)("XXX")$ for some constant $m$ $m$

Therefore a parabola with vertex at $(- 2, - 4)$ $(-2,-4)$ is of the form:
$XXX y = m {(x + 2)}^{2} - 4 XXX$ $color(white)("XXX")y=m(x+2)^2-4color(white)("XXX")$ for some constant $m$ $m$

If $(x, y) = (- 3, - 5)$ $(x,y)=(-3,-5)$ is a point on this parabola
$XXX - 5 = m {(- 3 + 2)}^{2} - 4$ $color(white)("XXX")-5 = m(-3+2)^2-4$

$XXX - 5 = m - 4$ $color(white)("XXX")-5 = m - 4$

$XXX m = - 1$ $color(white)("XXX")m=-1$

and the equation is $y = 1 {(x + 2)}^{2} - 4$ $y=1(x+2)^2-4$
graph{-(x+2)^2-4 [-6.57, 3.295, -7.36, -2.432]}

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