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#