Honors Pre-Calculus Unit 3 Review: Polynomial and Rational Functions

Complex numbers let you take square roots of negative numbers. The imaginary unit $i$ is defined by $i^2=-1$, so $\sqrt{-9}=3i$. A complex number has the form $a+bi$, with real part $a$ and imaginary part $b$. Every real number counts as complex with $b=0$.

To add or subtract, combine real parts with real parts and imaginary parts with imaginary parts, so $(3+2i)+(1-4i)=4-2i$. To multiply, distribute and then replace $i^2$ with $-1$. To divide, multiply the top and bottom by the conjugate of the denominator. The conjugate of $a+bi$ is $a-bi$, and $(a+bi)(a-bi)=a^2+b^2$ is always real.

Worked example. Simplify $\frac{3+2i}{1-4i}$. Multiply by $\frac{1+4i}{1+4i}$. The numerator becomes $3+12i+2i+8i^2=-5+14i$. The denominator becomes $1+16=17$. The answer is $-\frac{5}{17}+\frac{14}{17}i$.

Powers of $i$ repeat every four steps in the cycle $i,-1,-i,1$. Since 23 leaves a remainder of 3 when divided by 4, $i^{23}=i^3=-i$.

Complex numbers matter here because a quadratic with a negative discriminant still has two solutions, just not real ones. When the coefficients are real, those solutions come as a conjugate pair.

A quadratic function has degree 2, and its graph is a parabola. In standard form $f(x)=ax^2+bx+c$, the sign of $a$ sets the direction. When $a>0$ the parabola opens up and the vertex is a minimum. When $a<0$ it opens down and the vertex is a maximum. The y-intercept is $c$, and the axis of symmetry is $x=-\frac{b}{2a}$.

Vertex form $f(x)=a(x-h)^2+k$ shows the vertex $(h,k)$ directly. Watch the sign inside the parentheses, because $f(x)=2(x+3)^2-5$ has vertex $(-3,-5)$, not $(3,-5)$. You can convert standard form to vertex form by completing the square.

The discriminant $b^2-4ac$ tells you how many real x-intercepts to expect. A positive discriminant means two, zero means one (the vertex sits on the x-axis), and negative means none, with two complex roots instead.

The vertex is the highest or lowest point of the parabola, which makes quadratics a natural fit for optimization problems.

Worked example. You have 100 m of fencing for a rectangular garden. If the width is $x$, the length is $50-x$, so the area is $A(x)=x(50-x)=-x^2+50x$. The vertex is at $x=-\frac{50}{2(-1)}=25$. The best garden is 25 m by 25 m, with area $A(25)=625\text{ m}^2$.

A power function has a single term, $f(x)=kx^n$. A polynomial function adds terms like that together, $f(x)=a_nx^n+\dots+a_1x+a_0$, and every exponent must be a whole number (0, 1, 2, and so on). That means $f(x)=x^{1/2}$ and $f(x)=3x^{-2}$ are power functions but not polynomials.

The degree is the highest exponent, and the leading coefficient is the number in front of that term. Put the polynomial in descending order first, because the leading term isn't always written first. In $f(x)=5+2x-x^3$, the degree is 3 and the leading coefficient is $-1$.

End behavior describes where the graph heads as $x\to\infty$ and $x\to-\infty$. For large $|x|$, the leading term outgrows every other term, so it alone decides the ends. Even degree sends both ends the same direction, and odd degree sends them in opposite directions. A positive leading coefficient means the right end rises.

For example, $f(x)=-2x^4+7x^2-3$ has even degree and a negative leading coefficient, so both ends fall. As $x\to\pm\infty$, $f(x)\to-\infty$. The table below summarizes all four cases.

Polynomial graphs are smooth and continuous, with no breaks or sharp corners. You can sketch one from four pieces of information: end behavior, zeros, the multiplicity of each zero, and the y-intercept.

Find zeros by factoring completely and setting each factor equal to zero. Multiplicity is how many times a factor repeats. At a zero with odd multiplicity, the graph crosses the x-axis. At a zero with even multiplicity, it touches the axis and turns back. Higher multiplicity also flattens the graph near the zero.

A polynomial of degree $n$ has at most $n$ real zeros and at most $n-1$ turning points. So a graph with 4 turning points must come from a polynomial of degree 5 or higher.

When factoring doesn't work, use the Intermediate Value Theorem. Polynomials are continuous, so if $f(a)$ and $f(b)$ have opposite signs, a zero sits somewhere between $a$ and $b$.

Worked example. Sketch $f(x)=(x-3)^2(x+1)$. Multiplying the leading terms gives $x^3$, so the degree is 3 with a positive leading coefficient, and the graph falls on the left and rises on the right. The zero $x=-1$ has multiplicity 1, so the graph crosses there. The zero $x=3$ has multiplicity 2, so the graph touches and bounces. The y-intercept is $f(0)=(9)(1)=9$. Start low on the left, cross at $-1$, rise through $(0,9)$, turn down to touch at $3$, then rise to the right.

Dividing polynomials works like long division with numbers. For a nonzero divisor $d(x)$, the result fits the division algorithm $D(x)=d(x)\cdot q(x)+r(x)$. The remainder $r(x)$ is either zero or has a lower degree than $d(x)$. A remainder of 0 means $d(x)$ is a factor of $D(x)$.

Long division works for any divisor. Divide the leading terms, multiply, subtract, bring down the next term, and repeat. Subtraction is where most sign errors happen, so distribute the negative across every term you subtract.

Synthetic division is a shortcut for divisors of the form $x-c$, and it uses only coefficients. Two habits prevent most errors. Insert a 0 for every missing power, so $x^3+1$ becomes the coefficients $1,0,0,1$. Use the right sign for $c$, since dividing by $x+2$ means $c=-2$.

Worked example. Divide $2x^3-9x^2+12x-4$ by $x-3$. Use $c=3$ and the coefficients $2,-9,12,-4$. Bring down 2. Then $3\cdot2=6$, and $-9+6=-3$. Next $3\cdot(-3)=-9$, and $12-9=3$. Last $3\cdot3=9$, and $-4+9=5$. The quotient is $2x^2-3x+3$ with remainder 5, so $2x^3-9x^2+12x-4=(x-3)(2x^2-3x+3)+5$.

When you divide a polynomial of degree at least 1 by a linear polynomial, the quotient has degree one less than the dividend. If the divisor has degree 2 or higher, like $x^2+3$, use long division instead.

The Remainder Theorem says that dividing $P(x)$ by $x-c$ leaves a remainder of $P(c)$. The Factor Theorem follows directly. If $P(c)=0$, then $x-c$ is a factor, and the reverse is true too. Finding a zero and finding a factor are the same task.

The Rational Zero Theorem narrows your search. For a polynomial with integer coefficients and a nonzero constant term, any rational zero has the form $\pm\frac{p}{q}$, where $p$ divides the constant term and $q$ divides the leading coefficient. If the constant term is zero, factor out powers of $x$ first to find the zero at $x=0$. Test the candidates, divide out each zero you find, and repeat on the smaller polynomial. Once you reach a quadratic, factor it or use the quadratic formula.

The Fundamental Theorem of Algebra says a degree $n$ polynomial has exactly $n$ complex zeros, counting multiplicity. With real coefficients, nonreal zeros come in conjugate pairs, so if $2+i$ is a zero, so is $2-i$. That's why every cubic with real coefficients has at least one real zero. Descartes' Rule of Signs gives the maximum number of positive real zeros (sign changes in $P(x)$) and negative real zeros (sign changes in $P(-x)$).

Worked example. Find all zeros of $P(x)=x^3-3x^2-4x+12$. The candidates are $\pm1,\pm2,\pm3,\pm4,\pm6,\pm12$. Testing gives $P(2)=8-12-8+12=0$, so $x-2$ is a factor. Synthetic division leaves $x^2-x-6=(x-3)(x+2)$. The zeros are $2$, $3$, and $-2$.

A rational function is a ratio of two polynomials, $f(x)=\frac{p(x)}{q(x)}$. Its domain is every real number except where $q(x)=0$. Each excluded x-value turns into one of two features, and factoring tells you which one.

After canceling all common factors, check each excluded x-value. If the simplified denominator is nonzero there, the graph has a hole. If the simplified denominator is still zero there, the graph has a vertical asymptote, and the function heads toward $+\infty$ or $-\infty$ on each side.

Horizontal and slant asymptotes describe end behavior, and you find them by comparing degrees (see the table). A graph can cross a horizontal asymptote in the middle. A graph never crosses a vertical asymptote, because the function is undefined there.

The x-intercepts are the zeros of the simplified numerator that remain in the original domain. The y-intercept is $f(0)$, as long as 0 is in the domain.

Worked example. Analyze $f(x)=\frac{x^2-x-6}{x^2-4}=\frac{(x-3)(x+2)}{(x-2)(x+2)}$, where $x\ne-2$ and $x\ne2$. The factor $x+2$ cancels, leaving $\frac{x-3}{x-2}$. Plugging $-2$ into the simplified form gives $\frac{-5}{-4}=\frac{5}{4}$, so there is a hole at $(-2,\frac{5}{4})$. The vertical asymptote is $x=2$. The degrees are equal, so the horizontal asymptote is $y=\frac{1}{1}=1$. The x-intercept is $3$, and the y-intercept is $f(0)=\frac{-6}{-4}=\frac{3}{2}$.

An inverse function undoes the original. If $f(2)=7$, then $f^{-1}(7)=2$. On a graph, $f^{-1}$ is the reflection of $f$ across the line $y=x$, so the domain and range trade places. Keep in mind that $f^{-1}(x)$ is not the same as $\frac{1}{f(x)}$.

Only one-to-one functions have inverse functions. A function is one-to-one if it passes the horizontal line test, meaning no output repeats. The function $f(x)=x^2$ fails, since $f(3)=f(-3)=9$. Restrict it to $x\geq0$ and its inverse is $\sqrt{x}$. This is why radical functions are inverses of power functions. Odd powers like $x^3$ are already one-to-one, so $\sqrt[3]{x}$ works for all real numbers.

To find an inverse, write $y=f(x)$, swap $x$ and $y$, and solve for $y$. Check your answer with composition, since $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$ must both hold.

Worked example. Find the inverse of $f(x)=\sqrt{x-3}+1$. The domain is $x\geq3$ and the range is $y\geq1$. Swap to get $x=\sqrt{y-3}+1$. Then $x-1=\sqrt{y-3}$, so $(x-1)^2=y-3$ and $f^{-1}(x)=(x-1)^2+3$ for $x\geq1$. That restriction comes from the range of the original function.

When you solve a radical equation by squaring both sides, check every answer in the original equation. Squaring can create extraneous solutions.

Variation turns a sentence into an equation with a constant $k$, called the constant of variation. The process is the same every time. Write the general equation, plug in one known set of values to find $k$, then use the finished equation to answer the new question.

If $y$ varies directly with $x$, then $y=kx$. The graph is a line through the origin, and the ratio $\frac{y}{x}$ stays constant when $x\ne0$. If $y$ varies inversely with $x$, then $y=\frac{k}{x}$, so the product $xy$ stays constant and doubling $x$ halves $y$. Joint variation multiplies several variables, as in $z=kxy$. Combined variation mixes direct and inverse, as in $z=\frac{kx}{y}$.

Read the wording for powers. Varying directly with the square of $t$ means $d=kt^2$, and varying inversely with the square of $r$ means $I=\frac{k}{r^2}$. These are the power functions from earlier in the unit.

Worked example. Suppose an object is dropped from rest and air resistance is negligible. Its falling distance varies directly with the square of the time. It falls 64 ft in 2 s. So $64=k(2)^2$, which gives $k=16$ and $d=16t^2$. After 3 s it has fallen $16(3)^2=144$ ft.

A linear equation with a nonzero y-intercept, like $y=3x+2$, is not direct variation, because the ratio $\frac{y}{x}$ changes from point to point.