I put a picture this is the question
can an arrow or arrows be drawn from 10.3 so the relation in the diagram is a function? EXPLAIN YOUR ANSWER
will give brainliest:)

Yes, an arrow can be drawn from 10.3 so the relation is a function.

Step-by-step explanation:

Assuming the diagram on the left is the domain(the inputs) and the diagram on the right is the range(the outputs), yes, an arrow can be drawn from 10.3 and the relation will be a function.

The only time something isn't a function is if two different outputs had the same input. However, it's okay for two different inputs to have the same output.

In this problem, 10.3 is an input. If you drew an arrow from 10.3 to one of the values on the right, 10.3 would end up sharing an output with another input. This is allowed, and the relation would be classified as a function.

However, if you drew multiple arrows from 10.3 to different values on the right, then the relation would no longer be a function because 10.3, a single input, would have multiple outputs.

f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1

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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