Two linear functions are represented in different formats. Function 1:

x y
0 −2
2 0
3 1
5 3
Function 2:

Graph of a line passing through the point begin ordered pair negative 2 comma 0 end ordered pair and the point begin ordered pair 0 comma 4 end ordered pair.

Which statements are true?

Select each correct answer.

Function 1 has a greater rate of change than function 2.

Function 2 has a greater rate of change than function 1.

Function 1 has a greater y-intercept than function 2.

Function 2 has a greater y-intercept than function 1.

The statements that are true are;

Function 2 has a greater rate of change than function 1.

Function 2 has a greater y-intercept than function 1.

Step-by-step explanation:

Given that function 1 has the table;

x  y

0 -2

2  0

3   1

5   3

Finding the slope of the linear function gives the rate of change of the function. In this case,

m=Δy/Δx

m=3-1/5-3 = 2/2 =1

The equation of the linear function is given as;

m=Δy/Δx

y-3/x-5= 1

y-3=x-5

y=x-5+3

y=x-2

y-intercept is -2

In function 2, the line passes through points (-2,0) and (0,4)

Finding the slope of the line,

m₁=Δy/Δx

m₁=4-0/0--2

m₁= 4/2 =2

Rate of change is 2

Finding the equation of the line

y-4/x-0 = 2

y-4=2(x-0)

y-4 =2x

y=2x+4

The y-intercept is 4

The statements that are true are;

Function 2 has a greater rate of change than function 1.

Function 2 has a greater y-intercept than function 1.

the answer is c. 23/3

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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