AHL 2.8—Transformations of graphs, composite transformations

We need to find $f(x+4)$, which means replacing every $x$ in $f(x)=3x-2$ with $(x+4)$.

$f(x+4)=3(x+4)-2$

$3(x+4)-2=3x+12-2=3x+10$

The result is $f(x+4)=3x+10$.

We need to substitute $(x+4)$ into $f(x)=3x-2$ and simplify correctly.

$3x-6$ would result from $3(x-4)-2=3x-14$, or from incorrectly subtracting 4. Neither matches a valid substitution of $(x+4)$.

$3x+2$ would arise from $3x-2+4=3x+2$, which incorrectly adds 4 to the constant rather than substituting into the argument. This is wrong.

$3x+14$ would come from $3(x+4)+2$, incorrectly using $+2$ instead of $-2$. This is an arithmetic error.

The correct substitution gives $3(x+4)-2=3x+12-2=3x+10$.

$f(g(x))$ means substituting $g(x)$ into $f$. Since $g(x)=x+2$, we replace $x$ in $f$ with $(x+2)$.

$f(g(x))=f(x+2)=(x+2{)}^2-3$

$(x+2{)}^2-3=x^2+4x+4-3=x^2+4x+1$

Therefore $f(g(x))=x^2+4x+1$.

We expand $(x+2{)}^2-3$. A common error is failing to properly expand the square or mishandling the constant.

$x^2-1$ would arise from only squaring $x$ (getting $x^2$) and adding $4-3=1$, ignoring the cross-term $4x$. This is an incorrect expansion.

$x^2+4x-3$ comes from expanding $(x+2{)}^2$ correctly to $x^2+4x+4$ but then forgetting to subtract the $-3$, or incorrectly computing $4-3$ as $-3$.

$x^2+1$ would arise from $x^2-3+2=x^2-1$... no, or it might come from adding $g$ and $f$ incorrectly. Either way, this is not a valid composition result.

Correctly expanding $(x+2{)}^2-3=x^2+4x+4-3={\mathbf{x}}^2+4x+1$.