Using the completing-the-square method, rewrite f(x) = x2 + 4x − 1 in vertex form.

Answer

f(x) = (x + 2)² - 5

Explanation

f(x) = x² + 4x − 1

Let y = f(x)

y = x² + 4x − 1

y + 1 = x² + 4x

Make the right hand of the equation a perfect square by adding (b/2)² on both sides.

y + 1 + (4/2)² = x² + 4x + (4/2)²

y + 1 + 4 = x² + 4x + 2²

y + 5 = (x + 2)²

Subtracting 5 on both sides.

y + 5 - 5 = (x + 2)² - 5

y = (x + 2)² - 5

Now replace y with f(x)

f(x) = (x + 2)² - 5

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

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!