Math: Quadratics and Other Curves

This is Unit Eleven, part two, of ACT Preparation. Part one closed on the largest area a fence could enclose; this part widens from parabolas to polynomials, rational functions and variation. The habit carries over: let the form choose the method, and check the answer against the original and the situation. By the end you will be able to describe a polynomial's ends, zeros and turns. To write a polynomial from its graph. To use the remainder and factor theorems. To find a rational function's holes and asymptotes. And to solve variation problems.

Seven entries on polynomials, and on the theorems that tie zeros to factors.

How a polynomial's ends behave.

ACT 1019 End Behavior

A polynomial's end behavior follows its leading term. Even degree: both ends point the same way, up if the leading coefficient is positive. Odd degree: the ends point opposite ways. Negative x cubed plus 2x rises on the left and falls on the right. Only the leading term matters at the ends.

Zeros and their multiplicity.

ACT 1020 Zeros and Multiplicity

A factor x minus c raised to the n gives the zero c with multiplicity n. At odd multiplicity the graph crosses the x-axis; at even multiplicity it touches and turns. X minus one, squared, times x plus three touches at one and crosses at negative three.

Counting zeros and turns.

ACT 1021 Degree, Zeros and Turning Points

A polynomial of degree n has exactly n zeros, counting multiplicity and complex zeros, and at most n minus one turning points. A degree-four polynomial has four zeros and turns at most three times. Some zeros may be complex, so the graph can show fewer x-intercepts.

A polynomial from its graph, step by step.

ACT 1022 Reading Zeros From a Graph

Read each x-intercept, decide its multiplicity from crossing or touching, write the factors, and find the leading coefficient from one more point. A graph crossing at negative two, touching at one, and passing through zero, four is two times x plus two, times x minus one, squared: degree three, falling on the left and rising on the right. If the graph's ends disagree with the degree, a multiplicity is wrong.

The remainder theorem.

ACT 1023 The Remainder Theorem

When a polynomial is divided by x minus c, the remainder is f of c. X cubed minus 2x plus five divided by x minus two leaves f of two, nine. Division by x plus three uses negative three. Evaluating is often faster than dividing.

The factor theorem.

ACT 1024 The Factor Theorem

X minus c is a factor exactly when f of c is zero, so factors and zeros match. For x cubed minus 7x plus six, f of one is zero, so x minus one is a factor; f of three is twelve, so x minus three is not.

Using the factor theorem, step by step.

ACT 1025 Using the Factor Theorem

Identify c from the proposed factor, evaluate the polynomial at c keeping any unknown as a letter, and set the value to zero or read it as the remainder. For x plus two to be a factor of x cubed plus k x squared minus x plus two, evaluating at negative two gives 4k minus four, so k is one, and the check gives zero. A factor of x plus two means evaluating at negative two, not two.

Three entries on rational functions: holes, asymptotes, and the full analysis.

Holes and vertical asymptotes.

ACT 1026 Vertical Asymptotes and Holes

For a rational function, factor and cancel first. A factor that cancels leaves a hole at its zero; a zero of the reduced denominator gives a vertical asymptote. X minus two, over x minus two times x plus three, has a hole at two and an asymptote at negative three. Neither is in the domain.

Horizontal asymptotes.

ACT 1027 Horizontal Asymptotes

Compare degrees. A smaller numerator degree gives the asymptote y equals zero; equal degrees give the ratio of leading coefficients; a larger numerator degree gives none. 3x squared plus one over 2x squared minus five has asymptote three halves. A graph can cross its horizontal asymptote; it describes the ends only.

Analyzing a rational function, step by step.

ACT 1028 Analyzing a Rational Function

Factor and record the domain, cancel to find holes, read vertical asymptotes and x-intercepts from the reduced form, then compare degrees and find the y-intercept. 2x squared minus 2x over x squared minus one has a hole at one, an asymptote at negative one, an x-intercept at zero, and a horizontal asymptote at two. Find the hole's height from the reduced form: two x over x plus one at one is one, so the hole is at one, one.

Four entries on variation, and the whole unit in one problem.

Direct variation.

ACT 1029 Direct Variation

Y varies directly with x when y equals k x for a constant k: doubling x doubles y, and the graph is a line through the origin. Pay of $90 for six hours gives k equals fifteen, so ten hours pay $150. Find k first.

Inverse and joint variation.

ACT 1030 Inverse and Joint Variation

Y varies inversely with x when y equals k over x, so doubling x halves y; jointly with x and z when y equals k x z. Three hours at forty miles per hour gives k equals 120, so sixty miles per hour takes two hours. Inverse means the product stays constant.

A variation problem, step by step.

ACT 1031 Solving a Variation Problem

Translate the sentence into an equation with k, substitute the given values to find k, write the equation, then substitute the new values. In an invented model, a signal varies inversely with the square of the distance and is ninety units at two meters, so k is 360; at six meters it is 360 over thirty-six, ten units. Apply the square to the variable only: tripling the distance divides the strength by nine.

The whole unit in one problem.

ACT 1032 A Ball in the Air

This unit's tools describe curves and the equations that cut them, and the form of an expression chooses the method. In an invented model, a ball's height is negative 16t squared plus 64t plus eighty feet. The parabola opens down, so its vertex is the maximum: at two seconds, 144 feet. Setting the height to zero and dividing by negative sixteen gives t squared minus 4t minus five, which factors to five or negative one; time cannot be negative, so the ball lands at five seconds. The roots' sum, four, and product, negative five, match negative b over a and c over a, and the starting height, eighty feet, is the intercept. Discard a negative time as you would an extraneous root, and read whether a question asks for the input at the vertex or the value there.

Look at what you can now do. You can read a polynomial's ends from its leading term, and its crossings and touches from multiplicity. You can write a polynomial from its graph and check it against the ends. You can find a remainder or test a factor by evaluating at the right sign. You can find a rational function's holes and asymptotes, cancelling first. And you can solve variation problems by finding k first. Unit twelve turns to growth that multiplies instead of adding: exponentials and logarithms.