Identify the key features of these functions: f(x) = 3(x + 5)2 − 10

g(x) = 200(0.95)x

h(x) = -0.25(x + 3)(x − 1)

The vertex of f(x) is at
. The function g(x) is decreasing at a rate of
%. The zeros of h(x) are at
and
.

The vertex of f(x) is at (-5,-10).

The function g(x) is decreasing at a rate of 5%.

The zeros of h(x) are at x = -3 and x = 1.

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

It's vertex is the point

In which

If a<0, the vertex is a maximum point, that is, the maximum value happens at , and it's value is

Decaying exponential function:

A decaying exponential function has the following format:

In which A(0) is the initial quantity and r is the decay rate.

Zeros of a function:

Given a polynomial f(x), this polynomial has roots such that it can be written as: , in which a is the leading coefficient.

Vertex of f:

The function f is given by:

Expanding the calculations to place at the correct format:

Which means that

The x-value of the vertex is:

The y-value of the vertex is:

The vertex of f(x) is at (-5,-10).

Decreasing rate of g(x).

We have that:

1 - r = 0.95

So

r = 1 - 0.95 = 0.05

Which means that the decreasing rate is of 5%.

The zeros of h(x) are

The zeros are

x + 3 = 0 -> x = -3

x - 1 = 0 -> x = 1

So

The zeros of h(x) are at x = -3 and x = 1.

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!