How do you write a quadratic equation with Vertex: (-1,-3) x-intercept: 2?

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Answer 1 · 2016-05-25T07:17:17

$y= 1/3(x+1)^2-3$

The parabola is of the form:

$y= a( x- h)^2+k$

$"The vertex will be at the point" (\ h, \ k)$

$"Given the vertex at "(-1,-3)$

$h=-1" and " k=-3$

$"The " x "- intercept is at " 2 " so the point "(2,0)" is a particular point of the parabola"$

$0=a*(2-(-1))^2-3$

$0=a*(2+1)^2-3$

$0=a*(3)^2-3$

$9a-3=0$

$3a-1=0$

$a=1/3$

$"The quadratic equation is: "$

$y= 1/3(x+1)^2-3$