What is the vertex of the function f(x) = 4x2 + 3x + 3?

the vertex is at (h, k) => (-3/8, 39/16)

Step-by-step explanation:

One way of determining the vertex location is to "complete the square."

f(x) = 4x2 + 3x + 3 can be rewritten as    

f(x) = 4(x^2 + (3/4)x) + 3

We complete the square of (x^2 + (3/4)x ) as follows:

(x^2 + (3/4)x + 9/64 - 9/64)    or

       (x + 3/8)^2 - 9/64  

Now re-write f(x) = 4(x^2 + (3/4)x) + 3 (from above) as

                      f(x) = 4( (x + 3/8)^2 - 9/64 ) + 3, or

                       f(x) = 4(x + 3/8)^2 - 9/16 + 48/16, or

                       f(x) = 4(x + 3/8)^2 + 39/16

Comparing this to the standard vertex equation

                        f(x) = a(x - h)^2 + k, we see that h must be -3/8 and k must be 39/16.

Thus, the vertex is at (h, k) => (-3/8, 39/16).      

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!

step-by-step explanation: