How do you write a polynomial function for this graph?

first off, let's notice the graph touches the x-axis at -1 and 3, namely, those are the zeros/solutions/roots of the polynomial and therefore, the factors come from those points.

now, at -1, the graph doesn't cross the x-axis, instead it simply bounces off of it, that means the zero of x = -1, has an even multiplicity, could be 4 or 2 or 6, but let's go with 2.

at x = 3, the graph does cross the x-axis, meaning it has an odd multiplicity, could be 3 or 1, or 7 or 9, but let's use 1.

y = x³ - x² - 5x - 3

the graph has roots at x = - 1 and x = 3

there is a turning point at x = - 1 indicating a double root

given x = a and x = b are the roots of a polynomial then

(x - a) and (x - b ) are factors and the polynomial is the product of the factors

here x = - 1 ( double root ) ⇔ (x + 1)² is it's factors and

x = 3 hence (x + 3) is a factor

f(x) = (x + 1)²(x - 3)

     = (x² + 2x + 1)(x - 3)

     = x³ - x² - 5x - 3 ← is a possible polynomial

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1