Pls help me solve this!

The answer to your question is   y = 2/3 x + 2  

Step-by-step explanation:

Data

Equation = ?

slope = 2/3

P (-3, -1)

Process

1.- Identify x1 and y1

  x1 = -3

   y1 = -1

2.- Substitute the values in the equation of the point-slope form.

     y - y1 = m (x - x1)

-Substitution

     y - (-1) = 2/3(x - (-3))

-Simplification

     y + 1 = 2/3x + 3

- Solve for y

     y = 2/3 x + 3 - 1

-Simplify

     y = 2/3 x + 2  

Step-by-step explanation:

The point slope form is expressed as

y - y1 = m(x - x1)

Where

m represents the slope of the line.

The slope of the line is 2/3 and it passes through (- 3, - 1). To determine the equation, we would substitute x1 = - 3, y1 = - 1 and m= 2/3 into the point slope form equation. It becomes

y - - 1 = 2/3(x - - 3)

y + 1 = 2/3(x + 3)

The slope intercept form is expressed as

y = mx + c

Where c represents the y intercept

Therefore,

y + 1 = 2x/3 + 2

y = 2x/3 + 2 - 1

y = 2x/3 + 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! : )

cretdit to source

where are the graphs?

step-by-step explanation: