Consider the graph of the sixth-degree polynomial function f. Replace the values b, c, and d to write function f.
f(x)=(x-b)(x-c)^2(x-d)^3

f(x)=(x-1)(x+1)^2(x-4)^3

Step-by-step explanation:

The zeros of the function are x=-1, x=1, and x=4, so its factors are x+1, x-1, and x-4.

The graph touches, but doesn’t cross, the x-axis at x=-1. So the factor x+1 is a repeated factor with an even multiplicity.

The graph crosses the x-axis at x=4, but with a little curve before and after it meets this point. So the factor x+4 is a repeated factor with odd multiplicity.

The graph crosses the x-axis at x=1 with no curve before or after. So the factor x-1 is not a repeated factor.

Therefore, function f is defined by f(x)=(x-1)(x+1)^2(x-4)^3.

step-by-step explanation:

this is not a question

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!