Precalculus Solver with Step-by-Step Answers

• Domain and range, with each restriction traced to the thing that caused it — a denominator, an even root, or a logarithm.
• Composition and inverses, including the domain restriction an inverse inherits from the original function.
• Polynomial behaviour: zeros with multiplicity, end behaviour from the leading term, and the rational root theorem.
• Rational functions: vertical and horizontal asymptotes, slant asymptotes, and holes from cancelled factors.
• Exponential and logarithmic equations, using the rules that govern both, with the domain checked at the end.
• Sequences and series: arithmetic and geometric, sigma notation, partial sums, and infinite sums where the ratio allows.
• Conic sections: completing the square to reach standard form, then reading off centre, radius, foci, and asymptotes.
• Nonlinear systems, solved by substitution or elimination and checked against the graph.
• Transformations, and the order in which they must be applied.

Give the base of a logarithm explicitly. Written bare, $\log$ means base 10 in most school courses and base $e$ in several university ones, and the two produce different numbers from the same equation. Say which letter you are solving for when a formula carries several, and state the interval if a question about a sequence or a transformation only makes sense on part of the domain.

Every claim in that third analysis is testable with a calculator in under a minute. The horizontal asymptote at $y=2$ predicts that $f(100)$ sits near 2, and it does: $196/97=2.02$. The vertical asymptote at $x=3$ predicts a large output just beside it, and $f(3.01)=202$. The hole predicts nothing dramatic near $x=-2$ at all — values on both sides approach 1.6, which is the numerical difference between a hole and an asymptote.

Inverses have a different test, and it is composition rather than substitution. Feeding $f^{-1}(x)$ into $f$ should return $x$ for every legal input. One value is usually enough to expose a sign error, and it is faster than rereading the four lines of rearrangement that produced the inverse.

Log and exponential equations look varied and are not. Four situations cover nearly everything a precalculus course sets, and the right first move follows from where the unknown is sitting rather than from how complicated the expression looks.

Transformations are easier to trust once you have watched them happen. Change $a$ below and the parabola stretches vertically; change $c$ and the whole curve slides up or down without altering its shape. The vertex moves because $b$ shifts the axis of symmetry to $x=-b/2a$, which is worth confirming by eye before you memorise it.

One rule explains every transformation you will be asked about: changes made outside the function do what they look like, and changes made inside do the opposite. $f(x)+3$ lifts the graph three units, as expected. $f(x+3)$ shifts it three units left, because the function now reaches a given output at an $x$ three units earlier. The same asymmetry governs stretches: $2f(x)$ doubles the height, while $f(2x)$ halves the width. Derive the direction from that sentence each time and the table becomes unnecessary.

$\log (a+b)$ is not $\log a+\log b$. The product rule says $\log (ab)=\log a+\log b$; there is no rule for a sum inside the bracket at all. Test with numbers whenever you are unsure: ${\log }_{10}(10+10)\approx 1.30$, while ${\log }_{10}10+{\log }_{10}10=2$.

Condensing two logs into one enlarges the domain, so the condensed equation can have solutions the original does not. Write the domain restriction down before you start solving and treat it as part of the answer, not as an afterthought.

$f^{-1}(x)$ is the function that undoes $f$, not $1/f(x)$. The notation is genuinely unfortunate. The test is composition: if $f(f^{-1}(x))=x$ you have the inverse, and if you get something else you have a reciprocal.

A factor that cancels with the numerator produces a hole, not a vertical asymptote. The distinction matters because the function still has a finite limit there. Factor first, cancel second, and classify the leftovers third.

The thread running through all four is that precalculus notation packs a great deal of meaning into very little ink. A superscript −1, a base written below the line, a factor that cancels — each carries a condition the symbol itself does not display. Writing that condition down as it arises, rather than reconstructing it at the end, is most of what separates a clean solution from a merely plausible one, and it is the habit calculus assumes you turn up with.