Graph exponential and logarithmic functions, identify their asymptotes, use the reflection in the line y=x that makes them inverse functions, apply translations, reflections and dilations, and state the domain and range of each

What this dot point is asking

The previous two pages built the algebra of indices and logarithms; this page turns that algebra into pictures. NESA expects you to sketch the exponential curve $y = a^x$ and the logarithmic curve $y = \log_a x$, to know exactly where each one flattens out towards a line it never touches (its asymptote), and to use the single most important fact in the topic: the two curves are exact mirror images of each other in the line $y = x$, because they are inverse functions that undo one another. On top of the two basic shapes you must apply the usual graph transformations (translations up/down and left/right, reflections in each axis, and dilations) and, for every graph, state the domain and range. None of this needs calculus; it is all reading shape, asymptote and key points off a sketch. The pay-off is that the abstract statement "a logarithm is the index" finally becomes something you can see: the exponential turns an index into a number, the logarithm turns the number back into the index, and reflecting one curve in $y = x$ produces the other.

The answer

The shape of an exponential graph y = a^x

For a base $a > 1$, the graph of $y = a^x$ has one characteristic shape: it rises from left to right, getting steeper and steeper, and it stays entirely above the $x$-axis. Three features pin it down on any sketch, and all three come straight from the index laws:

• The $y$-intercept is $(0, 1)$, because $a^0 = 1$ for every base. Every exponential passes through this one point.
• At $x = 1$ the height is the base, the point $(1, a)$, because $a^1 = a$. This is what visibly separates $y = 2^x$ from $y = 3^x$: the steeper curve has the bigger base.
• The $x$-axis is a horizontal asymptote, the line $y = 0$. As $x \to -\infty$ the value $a^x$ becomes "as small as we like" but never reaches zero, because no power of a positive base is ever zero or negative.

The domain is all real $x$ (you can raise $a$ to any index) and the range is $y > 0$ (the output is always positive). The figure below shows $y = 2^x$ with these features; notice how the curve hugs the asymptote on the left without ever crossing it.

A base between $0$ and $1$ flips the picture left to right. For example $y = \left(\tfrac{1}{2}\right)^x$ can be rewritten $y = 2^{-x}$, which is $y = 2^x$ reflected in the $y$-axis: it decreases from left to right, still passing through $(0, 1)$ and still with asymptote $y = 0$ and range $y > 0$. This is the natural shape of decay (a halving each step), where the growth shape $y = 2^x$ is doubling.

The shape of a logarithmic graph y = log_a x

The graph of $y = \log_a x$ (with $a > 1$) is the exponential's twin. It rises from left to right but the other way round: it climbs steeply near the start and then flattens as $x$ grows. Its three features are the exponential's features with $x$ and $y$ swapped:

• The $x$-intercept is $(1, 0)$, because $\log_a 1 = 0$ for every base. Every logarithm passes through this one point.
• At $x = a$ the height is $1$, the point $(a, 1)$, because $\log_a a = 1$.
• The $y$-axis is a vertical asymptote, the line $x = 0$. As $x \to 0^+$ the value $\log_a x \to -\infty$, plunging downwards, because you are asking "what power of $a$ gives a number just above zero?" and the answer is a huge negative index.

The domain is $x > 0$ (only positive numbers have a logarithm) and the range is all real $y$. So the exponential's range $y > 0$ has become the logarithm's domain $x > 0$, and the exponential's domain (all reals) has become the logarithm's range, the swap that defines an inverse.

Why the two graphs are reflections in y = x

This is the central idea, and it is worth seeing clearly rather than memorising. The functions $y = a^x$ and $y = \log_a x$ are inverse functions: each undoes the other. Raising $a$ to a power and then taking $\log_a$ returns the original number, and vice versa, $\log_a(a^x) = x$ and $a^{\log_a x} = x$. Geometrically, undoing a function is reflecting its graph in the line $y = x$, because reflecting in $y = x$ swaps the horizontal and vertical coordinates: the point $(p, q)$ becomes $(q, p)$.

That swap is exactly the index-and-log relationship. A point $(p, q)$ on $y = a^x$ says $a^p = q$, that is, "$p$ is the index that gives $q$". Its mirror $(q, p)$ on $y = \log_a x$ says $\log_a q = p$, "the log of $q$ is $p$", the very same statement read backwards. So every point on one curve has its mirror on the other, and the whole logarithm graph is the whole exponential graph flipped across $y = x$. The table of values is even the same, with the columns swapped:

The diagram below draws both curves and the mirror line. The exponential's horizontal asymptote $y = 0$ reflects into the logarithm's vertical asymptote $x = 0$; its $y$-intercept $(0, 1)$ reflects into the logarithm's $x$-intercept $(1, 0)$. Hold a ruler along the dashed line and each curve folds exactly onto the other.

Transforming an exponential graph

Every transformation you met for parabolas and other curves works here unchanged. Start from a known exponential like $y = 2^x$ or $y = 3^x$ and apply the move; the key is to track the asymptote, the intercept and the domain/range as you go, because those are the marks.

• Translation up or down by $k$: $y = a^x + k$ shifts the curve $k$ units up (down if $k$ is negative). The asymptote moves with it, from $y = 0$ to $y = k$, and the range becomes $y > k$. The horizontal position is unchanged.
• Translation left or right by $h$: $y = a^{x - h}$ shifts the curve $h$ units right ($h$ negative shifts left). The asymptote stays $y = 0$ and the range stays $y > 0$, because moving sideways does not change the heights the curve reaches.
• Reflection in the $x$-axis: $y = -a^x$ flips the curve below the axis; the asymptote stays $y = 0$ but the range becomes $y < 0$.
• Reflection in the $y$-axis: $y = a^{-x}$ flips it left to right, turning growth into decay (this is the $0 < a < 1$ case again).
• Dilation (stretch) from the $x$-axis by factor $c$: $y = c\,a^x$ multiplies every height by $c$, so the $y$-intercept moves to $(0, c)$ while the asymptote stays $y = 0$.

The most-tested of these is the vertical shift, because it is the one that moves the asymptote. The figure shows $y = 3^x$ raised one unit to $y = 3^x + 1$: the whole curve lifts, the $y$-intercept goes from $(0, 1)$ to $(0, 2)$, and crucially the asymptote rises from $y = 0$ to $y = 1$.

One distinction trips up many students: $y = 2^x + 1$ and $y = 2^{x+1}$ are not the same. Adding $1$ to the output ($+1$ outside) shifts the curve up and moves the asymptote to $y = 1$; adding $1$ to the input ($+1$ in the exponent) shifts the curve left and leaves the asymptote at $y = 0$. They happen to share the $y$-intercept $(0, 2)$, but their asymptotes give them away. (Because $2^{x+1} = 2 \times 2^x$, the left shift is also exactly a vertical stretch by factor $2$, a tidy coincidence worth noticing.)

Transforming a logarithmic graph

Logarithms transform by the same rules, but watch the vertical asymptote, which is what a horizontal shift moves. Starting from $y = \log_a x$ (asymptote $x = 0$):

• Translation right or left by $h$: $y = \log_a(x - h)$ shifts the curve $h$ units right, and the vertical asymptote moves with it from $x = 0$ to $x = h$. The domain becomes $x > h$, because the number inside the log must still be positive, here $x - h > 0$.
• Translation up or down by $k$: $y = \log_a x + k$ lifts the curve $k$ units; the asymptote stays $x = 0$ and the domain stays $x > 0$.
• Reflections behave as before: $y = -\log_a x$ flips it in the $x$-axis, and $y = \log_a(-x)$ flips it in the $y$-axis (giving domain $x < 0$).

The figure shows $y = \log_2(x - 1)$, the curve $y = \log_2 x$ shifted right $1$ unit. The vertical asymptote moves to $x = 1$, the $x$-intercept moves from $(1, 0)$ to $(2, 0)$ (you need $x - 1 = 1$), and the domain is now $x > 1$.

Stating domain and range

For these graphs the domain and range are short to state once you know the shape, and the marks come from getting the strict inequality and the asymptote right.

• An exponential $y = a^x$ has domain all real $x$ and range $y > 0$. After a vertical shift $+k$, the range becomes $y > k$ (or $y < k$ if it is also reflected in the $x$-axis). The domain is still all real $x$, since a horizontal shift never restricts it.
• A logarithm $y = \log_a x$ has domain $x > 0$ and range all real $y$. After a horizontal shift, find the domain by demanding the inside of the log is positive: $y = \log_a(x - h)$ needs $x - h > 0$, so $x > h$. The range stays all real $y$.

The inequalities are strict ($>$, not $\ge$) because the curve approaches its asymptote without ever reaching it: the exponential never actually hits $y = 0$, and the logarithm is never actually defined at $x = 0$.

How exam questions ask about exponential and logarithmic graphs

The command words map onto specific features to show:

• "Sketch the graph of $y = \ldots$" means draw the right shape and label the key features: the intercept(s), one extra point such as $(1, a)$ or $(a, 1)$, and the asymptote drawn as a dashed line with its equation. A sketch with no asymptote labelled loses marks.
• "State the equation of the asymptote" is testing whether a shift moved it: vertical shift moves an exponential's $y = 0$ to $y = k$; horizontal shift moves a logarithm's $x = 0$ to $x = h$.
• "State the domain and range" wants the two short statements with strict inequalities, read off the asymptote.
• "Describe the transformation" wants the named move (translate up/down/left/right, reflect in the $x$- or $y$-axis, dilate) and the amount, with the reason ("$y$ replaced by $y - k$").
• "On the same axes, sketch $y = a^x$ and $y = \log_a x$" is asking you to use the reflection in $y = x$: draw the exponential, draw the dashed line $y = x$, then reflect to get the logarithm. Mark a mirror pair such as $(0, 1)$ and $(1, 0)$.
• "Hence" after a sketch usually means read a solution or a value off the graph you just drew (for example, where two curves meet).

This page builds directly on indices and index laws and on logarithms and their laws; if the reflection in $y = x$ feels slippery, re-reading "a logarithm is the index" is the surest fix, because the mirror is just that swap drawn out.