How do you write a quadratic equation with x-intercepts: -4,-2 ; point: (-6,8) ?

1 Answer
Jim G.
Oct 9, 2017

#y=x^2+6x+8#

Explanation:

#"given the x-intercepts (zeros ) of a quadratic"#

#"say "x=a" and "x=b#

#"then "(x-a)" and "(x-b)" are the factors"#

#"and the quadratic function is the product of the factors"#

#rArry=k(x-a)(x-b)#

#"where k is a multiplier which can be found if we are"#
#"given a point on the parabola"#

#"here "x =-4" and "x=-2#

#rArr(x+4)" and "(x+2)" are the factors"#

#rArry=k(x+4)(x+2)#

#"to find k substitute "(-6,8)" into the equation"#

#8=k(-2)(-4)=8krArrk=1#

#rArry=(x+4)(x+2)larrcolor(red)" in intercept form"#

#"distributing and simplifying gives"#

#y=x^2+6x+8larrcolor(red)" in standard form"#

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