if $f(x + 3) = 2x^2 - 3x - 1$ find $f(x + 1)$

Question

$f(x + 3) = (x + 3)(2x - 9) + 26$

Therefore,

$f(x + 1) = (x + 1)(2x - 9) + 26$

$f(x + 1) = 2x^2 - 7x + 17$

My steps are listed above. However in the answer key, it is stated that it will be $2x^2 - 11x + 13$ . They have substituted the $x$ in $f(x + 3)$ with $x - 2$ but I cannot understand why my approach is wrong.

Answer 1

You should change 2x - 9 too. Therefore,

$(+3)=(+3)(2−9)+26$,

$(+3)=(x+3)(2(x+3)-15)+26$,

$(+1)=(x+1)(2(x+1)-15)+26=(x+1)(2x-13)+26=2x^2−11+13$

Answer 2

You have that $f(x+3)=2x^2-3x-1$. Write $y=x+3$, then $x=y-3$ and thus $$f(y)=2(y-3)^2-3(y-3)-1=2y^2-15y+26.$$

Hence $f(x+1)=2(x+1)^2-15(x+1)+26=2x^2-11x+13$.

The key step is to first write $x+3$ as $y$ and change all of the $x$'s to $y-3$. In this way you rewrite the equation so that you get $f(y)$ in terms of $y$. Then you can replace $y$ by whatever you like.