Tom determines the system of equations below has two solutions, one of which is located at the vertex of the parabola.
equation 1: (x – 3)2 = y – 4
equation 2: y = -x + b
in order for this solution to be reasonable, which qualifications must be met?
b must equal 7 and a second solution to the system must be located at the point (2, 5).
b must equal 1 and a second solution to the system must be located at the point (4, 5).
b must equal 7 and a second solution to the system must be located at the point (1, 8).
b must equal 1 and a second solution to the system must be located at the point (3, 4).

Option A: b must equal 7 and a second solution to the system must be located at the point (2, 5)

Step-by-step explanation:

step 1

Find the vertex of the quadratic equation

The general equation of a vertical parabola in vertex form is  where  (h, k) is the vertex  we have  so , the vertex is the point (3,4)

step 2

Find out the value of b in the linear equation  we know that  if the vertex is a solution of the system of equations, then the vertex must satisfy both equations  substitute the value of x and the value of y of the vertex in the linear equation  for x=3, y=4.  

step 3

Find out the second solution of the system of equations  we have

> equation A

> equation B

Solve the system of equations by graphing . Remember that the solutions are the intersection points both graphs . The second solution of the system of equations is (2,5)

Therefore , b must equal 7 and a second solution to the system must be located at the point (2, 5)

the first dot plot best represents the data

plzplzplzplz mark brainliest and have a good day

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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