Construct a polynomial function with the stated properties. Third-degree, with zeros of - 3, - 1, and 2, and passes through the point (1, 7).

Using the Factor Theorem, the polynomial function is given by:

The Factor Theorem states that a polynomial function with roots is given by:

In which a is the leading coefficient.

In this problem, the zeros are -3, -1 and 2, hence .

The function is given by:

It passes through the point (1,7), which means that when , and this us used to find a.

Hence, the polynomial is:

A similar problem is given at link

Hello,

Third degree and zeroes are -3, -1 and 2 so we can write the function as below, a being a real

and we know that f(1)=7

so

a * 4 * 2 *(-1) = -8a=7

a = -7/8

Then

thanks

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!

y + 11 = (1/4)(x + 16)

step-by-step explanation: