Nonlinear functions

A function is a rule that turns an input into an output. $f(x)$ is not $f$ multiplied by $x$ — it is the output of the rule $f$ when the input is $x$. Everything in this topic follows from that.

To evaluate, substitute the input wherever $x$ appears and compute. To work out a composite such as $g(f(2))$, start inside the brackets: find $f(2)$ first, then feed that answer into $g$.

A shift changes where the graph sits. Adding inside the bracket, as in $f(x+2)$, moves the graph left by $2$ — which feels backwards until you see why: the function reaches any given output two units earlier.

Three questions below, one for each of those ideas.

The function $f$ is defined as $f(x) = 4x^2 + 2x - 1$.
What is the value of $f(-2)$?
Substitute $-2$ everywhere $x$ appears, and keep the brackets while you do it: $f(-2) = 4(-2)^2 + 2(-2) - 1$.
Work left to right. $(-2)^2 = 4$, so the first term is $4 \times 4 = 16$. The second is $2 \times -2 = -4$. So $f(-2) = 16 - 4 - 1 = 11$.
Two mistakes account for almost every wrong answer here. The first is squaring before applying the minus, reading $4(-2)^2$ as $(4 \times -2)^2 = 64$. The exponent belongs to the bracket only.
The second is dropping the sign and computing $4(2)^2 + 2(2) - 1 = 19$. A negative input carries its sign into every term.

The graph of $f(x)$ is shown above. No equation is given — everything you need is on the grid.
If $g(x) = f(x + 2)$, what is $g(1)$?
Do not try to identify the equation of the curve. Translate the question into a value you can read off the graph.
Since $g(x) = f(x+2)$, putting $x = 1$ gives $g(1) = f(1 + 2) = f(3)$.
So the question is really: how high is the curve at $x = 3$? Trace up from $3$ on the horizontal axis to the curve, then across to the vertical axis. It passes through $(3, 9)$, so $g(1) = 9$.
The common wrong answer is $1$, from reading the curve at $x = 1$ instead — that is $f(1)$, not $g(1)$. The other is $25$, from shifting the wrong way and reading $f(5)$.
Here is the idea worth keeping. $g$ is $f$ shifted left by $2$, so $g$ reaches at $x = 1$ whatever $f$ reached at $x = 3$. You never needed the equation of either curve.