Find a and b
2a +( b+i)i= 1+2i

a = 1

b = 2

Step-by-step explanation:

i is the imaginary number.

The sqroot(-1) = i

But useful for this problem is that i^2=-1

Also, for this problem we'll keep the right side of the equation just the way it is, change the left side to be in the same order (a plain number + an i term) Then we'll pick out the a and b that make it a true statement.

2a + (b + i)i = 1 + 2i

Distribute the i on the left (i acts just like a variable most of the time, even tho it isn't)

2a + bi + i^2 = 1 + 2i

Remember i^2 is -1.

2a + bi + -1 = 1 + 2i

2a + -1 + bi = 1 + 2i

(2a+-1) is the plain number.

bi is the i term.

2a+-1 = 1 and bi=2i

solve for a.

And b is 2.

2a + -1 +1 = 1+1

2a = 2

a = 1

c.

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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