What polynomial has a graph that passes through the given points? (-2, 2) (-1, -1) (1, 5) (3, 67) A. y = -x3 + 4x2 + 2x - 2 B. y = x3 + 4x2 + 2x - 2 C. y = x4 + 4x3 + 2x2 - 2x D. y = x3 - 4x2 - 2x - 2
test them around say test A) for the 1st point y = -x^3 + 4x^2 + 2x - 2 using (-2, 2) \(\bf (2) = -(-2)^3 + 4(-2)^2 + 2(-2) - 2\)
so you get 2 = 8+16-4-2 or 2 = 18 of courses 2 is not equals 18, so it's not A obviously
so check B
so try b and it would be (2)= (-2)2 +4(-2)2+ 2(-2) -2
which gives you 2 =-6?
I think
\(\bf y = x^3 + 4x^2 + 2x - 2 \\ (-2, 2)\\ (2) = (-2)^3 + 4(-2)^2 + 2(-2) - 2\)
-8 + 16 - 4 -2 2
-2 * -2 * -2 -2 * -2 = +4 +4 * -2 = -8
i got my signs mixed up.
\(\bf y = x^3 + 4x^2 + 2x - 2 \\ (-1, -1)\\ (-1) = (-1)^3 + 4(-1)^2 + 2(-1) - 2\)
2= -1 no dice on the 2nd point
so c
dunno let's test it
okay
\(\bf y = x^4 + 4x^3 + 2x^2 - 2x \\ (-2, 2)\\ (2) = (-2)^4 + 4(-2)^3 + 2(-2)^2 - 2(-2)\)
\[2= -2(4)+4(-2)(3)+ 2(-2) 2-2(-2)\]
okay let me solve real quick
2 = 16-32+8+4 2 = -4 no dice
i got the same. alright lets try d
so hehehe, by elimination there's only one candidate left hehe When you have eliminated the impossible, whatever remains, however improbable, must be the truth. ~~ Sherlock Holmes, The Sign of Four Mistakes are always initial. ~
well i tried d and got 2=-6
\(\bf y = x^3 - 4x^2 - 2x - 2 \\ (-2, 2)\\ (2) = (-2)^3 - 4(-2)^2 - 2(-2) - 2\)
hmmm
i think it is b
i did the other numbers and i got it
hmmm lemme check B then, thus far the only one that gave a good one
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