anonymous
  • anonymous
Find a quadratic model for the set of values: (-2, -20), (0, -4), (4, -20). Show your work
Algebra
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
sdfgsdfgs
  • sdfgsdfgs
Since u need to show ur work....the general quadratic model is y=ax^2 + bx + c Now we need to find the constants a, b, c ok?
anonymous
  • anonymous
How do we do that
sdfgsdfgs
  • sdfgsdfgs
U are given 3 points: (-2, -20), (0, -4), (4, -20) So u can use them to solve for abc

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I suck at this and I don't get it
sdfgsdfgs
  • sdfgsdfgs
I will do the first one for u, @Raylynngrace :) y=ax^2 + bx + c for (-2, -20) so -20 = a*(-2)^2 + b*(-2) + c -20 = 4a - 2b + c ok?
anonymous
  • anonymous
Okay I get that now lol
sdfgsdfgs
  • sdfgsdfgs
Cool! Show me how u set up with the other two given points plz?
anonymous
  • anonymous
idk if I did this right lol but for the next one I got y=ax^2+bx+c -4=a*(0)^2+b*(0) -4=a-b+c
sdfgsdfgs
  • sdfgsdfgs
good try but u missed something...it should be like -4=a*(0)^2+b*(0) + c -4=0+0+c c=-4 hey we found c already!
anonymous
  • anonymous
Least I tried lol can you help me with the last one
sdfgsdfgs
  • sdfgsdfgs
nope u do it plz? i will help of cus :)
anonymous
  • anonymous
y=ax^2+bx+c -20=a*(4)^2+b*(4) -20=16a+4b
anonymous
  • anonymous
Did I get this one wrong to @sdfgsdfgs
sdfgsdfgs
  • sdfgsdfgs
@raylynngrace hahahaa u keep forgetting the c term....so it should be: -20=a*(4)^2+b*(4)+c -20=16a+4b+c now u got c=-4 already and from the first point: -20 = 4a - 2b + c U can solve for a and b :)
anonymous
  • anonymous
@Raylynngrace This will help you a bit. Try !! http://www.softschools.com/math/algebra/quadratic_functions/quadratic_function_with_three_points/
anonymous
  • anonymous
Thank you both for your help @sdfgsdfgs @JoshKoikkara

Looking for something else?

Not the answer you are looking for? Search for more explanations.