Find polynomial function p(x) such that p has the degree and p(x) = 0 has the roots listed

degree of p= 2
zeros= 1 + 6i

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

idk

step-by-step explanation:

i use my brain for prombles like this maybe you could try it

step-by-step explanation: