How do you write #f(x)=2x^2+12x+12# in vertex form?

Question

Ethan Adkins · Accepted Answer

#y=2(x+3)^2-6#

Vertex form is also known as completing the square

#color(blue)("Step 1")#

Write as:

#y=2(x^2+6x)+12+k#

When we start to change things the equation becomes untrue. So we need to introduce the correction #k# to compensate for this. The value of #k# is calculated at the end.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

Move the power of 2 from #x^2# to outside the brackets.

#y=2(x+6x)^2+12+k#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Halve the 6 from #6x#

#y=2(x+3x)^2+12+k#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Remove the #x# from #3x#

#y=2(x+3)^2+12+k# .....................Equation(1)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

Determine the value of #k#

If you were to square the bracket we would have #color(magenta)(3^2)# form #color(green)(2)(x+ color(magenta)(3))^2#. Also this is multiplied by the #color(green)(2)# from outside the bracket giving:

#color(green)(2xx)color(magenta)(3^2) larr" this is the error we introduced"#

So we write:

#color(green)(2xx)color(magenta)(3^2)+k=0#

#=> 18+k=0 -> k=-18#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5 - The final equation")#

Substitute this into equation(1)

#y=2(x+3)^2+12-18#

#y=2(x+3)^2-6#

#color(red)("I have superimposed both graphs so that you can see they are the same.")#

Kennedy Adams · Answer

undefined To write the quadratic function $f(x) = 2x^2 + 12x + 12$ in vertex form, follow these steps:

1. Factor out the leading coefficient $2$ from the terms involving $x^2$ and $x$:

$$f(x) = 2(x^2 + 6x) + 12$$

2. Complete the square inside the parentheses by adding and subtracting the square of half the coefficient of $x$, which is $(6/2)^2 = 9$:

$$f(x) = 2(x^2 + 6x + 9 - 9) + 12$$

3. Rewrite the expression inside the parentheses as a perfect square trinomial:

$$f(x) = 2[(x + 3)^2 - 9] + 12$$

4. Distribute $2$ and simplify:

$$f(x) = 2(x + 3)^2 - 18 + 12$$

5. Combine the constant terms:

$$f(x) = 2(x + 3)^2 - 6$$

So, the vertex form of $f(x) = 2x^2 + 12x + 12$ is $f(x) = 2(x + 3)^2 - 6$.