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Find an equation for the quadratic that passes through the points (−1,8), (2,−4), and (−6,−12).

 Jun 8, 2021
 #1
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y=ax^2+bx+c

first substitute (-1, 8) (1 for x and 8 for y)

8=a-b+c

then substitute (2, -4) (2 for x and -4 for y)

-4=4a+2b+c

then substitute (-6, -12) (-6 for x and -12 for y)

-12=36a-6b+c

now we have these three equations:

8=a-b+c

-4=4a+2b+c

-12=36x-6x+c

so a=-1

b=-3

and c=6

and the quadratic equation's equation is y=-x^2-3x+6

 

JP

 Jun 8, 2021

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