we
know that if we have a polynomial with real coefients and one root is
a+bi, another root is a-bi
so
some
roots are
3,-4,6+5i,
6-5i
for
roots, r1 and r2
teh
factors are
(x-r1)(x-r2)
so
we
can put them in and
(x-3)(x+4)(x-6-5i)(x-6+5i)
(x^2+x-12)(2x^2-12x+61)
x^4-11x^3+37x^2+205x-732
to
keep the same coefients, we multiply whole thing by c
c(x^4-11x^3+37x^2+205x-732)
we
don't know what c is
we
know that
f(1)=250
f(1)=c(x^4-11x^3+37x^2+205x-732)
find
c
f(1)=c(1^4-11(1^3)+37(1^2)+205(1)-732)=250
f(1)=c(-500)=250
-500c=250
divide
both sides by -500
c=-1/2
the
polynomial (factored with real coefients) is
f(x)=-0.5(x-3)(x+4)(2x^2-12x+61)
the angles are p and q im pretty sure
So your equation is in the form (y-k)^2 = 4p(x-h), where focus is at (h+p,k) and vertex is at (h,k) If h = 4 from the vertex and h + p = 3 from the focus, p = -1 in your parabola. So your equation looks like (y--7)^2 = -4(x - 4). Clean that up and you're all set.
Answer:24 beads 3x8=y y=24
Step-by-step explanation:
