Exam Graph Drill

Identity $f(x) = x$: straight line through the origin, slope 1. Domain $(-\infty, \infty)$, range $(-\infty, \infty)$. Plot $(-2,-2)$, $(0,0)$, $(2,2)$ and join.

Read $y = -(x-2)^2 + 3$ step by step: start from $x^2$, move right 2, flip (the minus), move up 3. Vertex $(2, 3)$, opens down, y-intercept $-(0-2)^2 + 3 = -1$.

Values: $f(-1) = -1 + 2 = 1$, $f(0) = 0^2 = 0$ (the closed dot, not the open one at 2), $f(2) = 4$, $f(3) = 1$. Range: $(-\infty, 2) \cup [0, 4] \cup \{1\} = (-\infty, 4]$.

Graph $2x + 3y = 12$. Put $x = 0$: $3y = 12$, $y = 4$. Put $y = 0$: $2x = 12$, $x = 6$. Plot $(0, 4)$ and $(6, 0)$ and join. Slope $= -\tfrac{2}{3}$: down 2 for every 3 across.

Finished sketch: five labelled points (vertex, two x-intercepts, y-intercept and its mirror), the dashed axis, a smooth symmetric U. Vertex form: $f(x) = (x - 2)^2 - 9$.

Left of 200 the cost line is above the revenue line: loss (C − R). Right of 200 revenue is above: profit $P(x) = 20x - 4000$. At 300 units: $P = 2000$ TL, the vertical gap between the lines.

$R = C$: $100x - 2x^2 = 20x + 600 \Rightarrow 2x^2 - 80x + 600 = 0 \Rightarrow x^2 - 40x + 300 = 0 \Rightarrow (x - 10)(x - 30) = 0$. Break-even at $x = 10$ and $x = 30$. Profit only between them.

$y = 2^x$ ($b \gt 1$): always passes through $(0, 1)$, rises left to right, never touches the x-axis. Horizontal asymptote $y = 0$ (as $x \to -\infty$). Domain all reals, range $(0, \infty)$.

$y = \ln x$: through $(1, 0)$ and $(e, 1)$, rises slowly, plunges to $-\infty$ as $x \to 0^+$. Vertical asymptote $x = 0$. Domain $(0, \infty)$, range all reals. $\ln x \lt 0$ for $0 \lt x \lt 1$.

A typical exam graph. Open dot: the curve comes arbitrarily close, but that point is not on the graph. Closed dot: the actual value. Limits come from the curve's height near $a$; the value $g(a)$ comes from the closed dot.

$\lim_{x\to2^+} \frac{1}{x-2} = +\infty$ (tiny positive denominator), $\lim_{x\to2^-} \frac{1}{x-2} = -\infty$ (tiny negative). Vertical asymptote $x = 2$; two-sided limit does not exist. As $x \to \pm\infty$, $y \to 0$: horizontal asymptote $y = 0$.

Removable (hole): $f(x) = \frac{x^2-4}{x-2}$. $f(2)$ is undefined, but $\lim_{x\to2} f(x) = 4$. Fails condition 1. Defining $f(2) = 4$ would remove it.

Make the pieces meet: $3(2) + k = 2^2 + 1 \Rightarrow 6 + k = 5 \Rightarrow k = -1$. Left limit = right limit = $f(2) = 5$: continuous.

$f(x) \gt 0$ on $(-3, 1) \cup (4, \infty)$. Open circles: $-3$ and $4$ give $f = 0$ (not $\gt 0$), and $1$ is never allowed. For $\ge 0$ you would close $-3$ and $4$, but $1$ stays open.