The graphed line shown below is y = negative 3 x + 6. On a coordinate plane, a line goes through points (1, 3) and (2, 0).

Which equation, when graphed with the given equation, will form a system that has no solution?
y = 3 x + 6
y = negative 3 (x + 6)
y = negative 3 (x minus 2)
y = 3 (x minus 2)

answer: 0+0=0

step-by-step explanation:

that is the only answer that can make 0

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! : )

cretdit to source

lol.

it's ok we all do this sometimes.

-aparri

answer: it actually depends on which is the independent variable and which is the dependent variable. let us say we're looking at a linear equation. y=2x+3. we theoretically could switch the equation to be x=(y-3)/2 and it would be the same line. so either a causes b or it can be that b causes a. so technically the answer is false

step-by-step explanation: