\(f(x+2)=4x^2-2\)

So the question is to find the parent function f(x)

I found that the answer is \(f(x)=4x^2-16x+14\)

but I only got that by a long game of trial and error :(

To get started I did the following:

Factor: \(4x^2-2=2(x^2-1)=2(x-1)(x+1)\)

Plug in the value of f "x+2": \(2([x+2]-1)([x+2]+1)=2(x+1)(x+3)\)

Simplify: \((2x+2)(x+3)=2x^2+6x+2x+6=2x^2+8x+6\)

The result is something close... my first instinct was to double coefficents which gets really close to the answer but where does the negative come from in the answer on the second term and why is the last term 14 and not 12?

MY QUESTION is what is a standard non-guessing method of solving for the parent function?

Try to be genneral too so I can solve functions of the like : \(2 f(-x^2), 3f^2(\sqrt{x}/5),etc\)

Guest Dec 31, 2017

#1**+2 **

f(x + 2) = 4x^2 - 2

Notice that this must be the parent function, f(x), shifted to the left by 2 units.....thus....we can shift the function 2 units to the right to find the parent function....so we have

4(x - 2)^2 - 2 =

4(x^2 - 4x + 4) - 2 =

4x^2 - 16x + 16 - 2 =

4x^2 - 16x + 14 = f(x)

CPhill Jan 1, 2018

#2**+3 **

You should begin by deducing what the 'parent function' looks like, and then working from left to right, rather than right to left as you have tried to do.

For this example, it should be clear that the 'pf ' will be a quadratic, so start with the assumption that

\(\displaystyle f(x) = ax^{2}+bx +c\).

In that case,

\(\displaystyle f(x+2)=a(x+2)^{2}+b(x+2)+c\\ =ax^{2}+(4a+b)x+(4a+2b+c)\\\equiv 4x^{2}-2 .\)

Equating coefficients across the identity leads to, (in turn), a = 4, b = -16 and c = 14.

If though, for example, you were given \(\displaystyle f(x^{2}+2)=4x^{2}-2\), then your presumed 'pf ' would be \(\displaystyle f(x)=ax+b\), leading to \(\displaystyle f(x^{2}+2)=a(x^{2}+2)+b\) , etc. .

Tiggsy

Guest Jan 1, 2018