Graphs of elementary functions

Constant function. For $f(x) = c$, every input gives the same output, so the graph is a horizontal line sitting at height $c$. There is no restriction on which $x$ you can plug in, so the domain is all of $\mathbb{R}$, but only one $y$-value ever comes out, so the range is the single-point set $\{c\}$.

Identity function. For $f(x) = x$, the output always equals the input, so the graph is the straight line through the origin tilted at exactly $45°$. Here both domain and range are the whole of $\mathbb{R}$ — nothing is excluded on either axis.

Polynomial functions. A polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ only ever adds, subtracts, and multiplies its input — operations that never fail for any real number — so its domain is always $\mathbb{R}$, regardless of degree. The shape of the graph, though, depends heavily on the degree: A linear function $f(x) = mx + c$ draws a straight line of slope $m$; its range is $\mathbb{R}$ unless $m = 0$, which reduces it to the constant case above. A quadratic function $f(x) = ax^2 + bx + c$ draws a parabola, opening upward when $a > 0$ and downward when $a < 0$. Its turning point, the vertex, sits at

where: $a$ and $b$ are the quadratic's leading and linear coefficients. Substituting this $x$ back into $f$ gives the vertex's $y$-coordinate, which is the minimum of the range if $a>0$ or the maximum if $a<0$ — this single point fixes the range of the entire parabola. Higher-degree polynomials produce smooth, unbroken curves with up to $n-1$ turning points, but at Class 11 level the expectation is only to confirm the domain is $\mathbb{R}$ and to read the range qualitatively from the sketch's overall shape.

Rational functions. A rational function $f(x) = \dfrac{p(x)}{q(x)}$ is a ratio of two polynomials, and division breaks down wherever the denominator is zero — so the domain excludes every root of $q(x)$. The simplest case, $f(x) = \dfrac{1}{x}$, has domain $\mathbb{R} - \{0\}$; its graph is a hyperbola sitting in the first and third quadrants, curving in toward both axes but never touching them. Those two lines the curve approaches but never reaches — here $x=0$ and $y=0$ — are called asymptotes, and its range is $\mathbb{R} - \{0\}$ for the same reason: the equation $\dfrac{1}{x}=0$ has no solution.

Problem. Sketch $f(x) = x^2-4x+3$ and $g(x) = \dfrac{2}{x-1}$, and state the domain and range of each. (This reuses the two functions from the Pass-1 article, now solved back to back to compare a polynomial and a rational function directly.)

Step 1. For $f(x)=x^2-4x+3$: since $a=1>0$, the parabola opens upward. Its vertex is at $x=-\frac{-4}{2(1)}=2$, and $f(2)=4-8+3=-1$, so the vertex is $(2,-1)$.

Step 2. Factor to find the $x$-intercepts: $x^2-4x+3=(x-1)(x-3)$, giving roots $x=1$ and $x=3$; the $y$-intercept is $f(0)=3$. Plotting the vertex and these three points and drawing the symmetric upward curve through them gives the sketch.

Step 3. Since $f$ is a polynomial, its domain is $\mathbb{R}$; since the vertex $(2,-1)$ is the lowest point on an upward parabola, every $y \geq -1$ is attained, so $\text{Range}(f) = [-1,\infty)$.

Step 4. For $g(x)=\dfrac{2}{x-1}$: the denominator is zero at $x=1$, so $\text{Domain}(g)=\mathbb{R}-\{1\}$, and $x=1$ is a vertical asymptote. As $x\to\pm\infty$, $g(x)\to 0$, so $y=0$ is a horizontal asymptote — and since $g(x)=0$ has no solution, $\text{Range}(g)=\mathbb{R}-\{0\}$.

Step 5. Plotting a few guide points — $g(2)=2$, $g(0)=-2$, $g(3)=1$, $g(-1)=-1$ — shows two separate branches, one for $x>1$ and one for $x<1$, each hugging the asymptotes without touching them.

Final answer. $f$: domain $\mathbb{R}$, range $[-1,\infty)$, vertex $(2,-1)$. $g$: domain $\mathbb{R}-\{1\}$, range $\mathbb{R}-\{0\}$, asymptotes $x=1$ and $y=0$.

Competitive stretch

For $h(x) = \dfrac{3}{x+2} - 1$, the same reasoning gives domain $\mathbb{R}-\{-2\}$ (denominator zero at $x=-2$) and, since $h(x)=-1$ can never be reached (it would need $\frac{3}{x+2}=0$), range $\mathbb{R}-\{-1\}$ — the vertical asymptote shifts to $x=-2$ and the horizontal asymptote shifts to $y=-1$, exactly tracking the shift added to $\frac{1}{x}$.

• Assuming a polynomial has some domain restriction — it never does; domain $\mathbb{R}$ holds for every polynomial regardless of degree.
• Sketching a parabola opening the wrong way — always read the sign of $a$ first: positive means upward, negative means downward.
• Drawing a rational function's two branches as one continuous curve across the vertical asymptote — the branches on either side never connect.
• Reading a parabola's range off the $y$-intercept instead of the vertex — only the vertex's $y$-value bounds the range.

Ask the tutor: Give me a fresh quadratic or rational function to sketch and check my vertex and asymptotes.