The graph of a quadratic function is a transformation of the graph of y=x2.
The graph of a quadratic function is a parabola with vertex
(h,k)=(−b2a,f(−b2a)).
The maximum (if a<0) or minimum (if a>0) value of a quadratic function f(x)=ax2+bx+c is the y-coordinate, k, of the vertex of the parabola.
Sketch the graph off(x)=−2x2+4x+6.
Solution
We can put f into standard form:
f(x)=−2x2+4x+6=−2(x−1)2+8.
The graph of f is the graph of y=x2 shifted 1 unit to the right, stretched vertically by a factor of 2, reflected about the x-axis, and shifted vertically up 8 units.
The vertex is (−42(−2), f (−42(−2)))=(1,f(1))=(1,8).
The graph opens downward because a=−2<0. The x-intercepts are −1 and 3, the solutions of the equation −2x2+4x+6=0. The domain is (−∞,∞) and the range is (−∞,8]. The function f is increasing on (−∞,1) and decreasing on (1,∞). The maximum value of f is 8.
3.2 Polynomial Functions
A function f of the form f(x)=anxn+an−1xn−1+…a2x2+a1x+a0,an≠0, is a polynomial function of degree n.
The graph of a polynomial is smooth and continuous.
The end behavior of the graph of a polynomial function depends on the sign of the leading coefficient and the degree (even-odd) of the polynomial.
A real number c is a zero of a function f if f(c)=0. Geometrically, c is an x-intercept of the graph of y=f(x).
If in the factorization of a polynomial function f the factor (x−a) occurs exactly m times, then a is a zero of multiplicitym. If m is odd, the graph of y=f(x) crosses the x-axis at a; if m is even, the graph touches but does not cross the x-axis at a.
If the degree of a polynomial function f is n, then f has, at most, n real zeros and the graph of f has, at most, n−1 turning points.
Intermediate Value Theorem. Let f be a polynomial function and a and b be two numbers such that a<b. If f(a) and f(b) have opposite signs, then there is at least one number c, with a<c<b, for which f(c)=0.
See Example9 on page 342 for the procedure for graphing a polynomial function.
Describe the end behavior off(x)=3x4−5x3+7x2−4x+9.
Solution
The function f(x)=3x4−5x3+7x2−4x+9 is a polynomial function of degree 4.
Based on the fact that the leading coefficient, 3, is positive and that the degree, 4, is even, we know that the end behavior of f is similar to that of y=x4, that is:
f(x)→∞asx→−∞andf(x)→∞asx→∞.
Sketch the graph off(x)=x3−x2.
Solution
The polynomial f(x)=x3−x2 can be factored as f(x)=x3−x2=x2(x−1). The zeros of f are x=0 and x=1, the solutions of the equation f(x)=0. The zero, 0, has multiplicity 2 because the factor (x−0) occurs exactly twice. The graph touches but does not cross the x-axis at 0. The zero, 1, has multiplicity 1 because the factor (x−1) occurs exactly 1 time. The graph crosses the x-axis at 1. The degree of f is 3, so f has, at most, 3 real zeros and the graph of f has, at most, 2(=3−1) turning points.
Refer to Example9 on page 342 to review sketching the graph of an equation by plotting points.
3.3 Dividing Polynomials
Division Algorithm. If a polynomial F(x) is divided by a polynomial D(x)≠0, there are unique polynomials Q(x) and R(x) such that F(x)=D(x)Q(x)+R(x), where either R(x)=0 or deg R(x)<deg D(x). In words, “The dividend equals the product of the divisor and the quotient plus the remainder.”
Synthetic division is a shortcut for dividing a polynomial F(x) by x−a.
Remainder Theorem. If a polynomial F(x) is divided by x−a, the remainder is F(a).
Factor Theorem. A polynomial function F(x) has x−a as a factor if and only if F(a)=0.
Use synthetic division to divideF(x)=3x3+8x2+5x−4byx+2.
If F(x)=x2−3x−10, then F(5)=52−3(5)−10=25−15−10=0. By the Factor Theorem (x−5) is a factor of F. Because x−3 is a factor of G(x)=x2−9,G(3)=0.
3.4 The Real Zeros of a Polynomial Function
Rational Zeros Theorem. If pq is a rational zero in lowest terms for a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.
Descartes’s Rule of Signs. Let F be a polynomial function with real coefficients.
The number of positive zeros of F is equal to the number of variations of signs of F(x) or is less than that number by an even integer.
The number of negative zeros of F is equal to the number of variations of sign of F(−x) or is less than that number by an even integer.
Rules for Bounds on the Zeros. Suppose a polynomial F is synthetically divided by x−k.
If k>0 and each number in the last row is zero or positive, then k is an upper bound on the zeros of F.
If k<0 and the numbers in the last row alternate in sign, then k is a lower bound on the zeros of F.
Find all zeros off(x)=2x3−x2−9x−4.
Solution
The only rational numbers that can possibly be zeros of f(x)=2x3−x2−9x−4 are ±1,±12,±2, and ±4. From
we see that −12 is a rational zero.
So, f(x)=(x+12)(2x2−2x−8). Because 2x2−2x−8 has no rational zeros, −12 is the only rational zero for f. We can use the quadratic equation to solve the depressed equation 2x2−2x−8=0, to get x=1±17−−√2. The zeros of f are: {−12,1−17−−√2,1+17−−√2}.
3.5 The Complex Zeros of a Polynomial Function
Fundamental Theorem of Algebra. An nth-degree polynomial equation has at least one complex zero.
Factorization Theorem for Polynomials. If P(x) is a polynomial of degree n≥1, it can be factored into n (not necessarily distinct) linear factors in the form P(x)=a(x−r1)(x−r2)…(x−rn), where a,r1,r2,…,rn are complex numbers.
Number of Zeros Theorem. A polynomial of degree n has exactly n complex zeros, provided a zero of multiplicity k is counted k times.
Conjugate Pairs Theorem. If a+bi is a zero of the polynomial function P (with real coefficients), then a−bi is also a zero of P.
Find all the zeros off(x)=x3−6x2+13x−10,given that2+iis one of its zeros.
Solution
The polynomial f(x)=x3−6x2+13x−10 has at least one and at most three real zeros.
Because 2+i is a zero of f(x)=x3−6x2+13x−10, its conjugate 2−i is also a zero of f.
By the factorization theorem the product (x−[2+i])(x−[2−i])=x2−4x+5 is a factor of f(x). Dividing f(x) by x2−4x+5, we get the quotient: x−2. So, f(x)=(x−2)(x2−4x+5). The three zeros of f are: {2,2−i,2+i}.
3.6 Rational Functions
A function f(x)=N(x)D(x), where N(x) and D(x) are polynomials and D(x)≠0, is called a rational function. The domain of f is the set of all real numbers except the real zeros of D(x).
The line x=a is a vertical asymptote of the graph of f if |f(x)|→∞ as x→a+ or as x→a−.
If N(x)D(x) is in lowest terms, then the graph of f has vertical asymptotes at the real zeros of D(x).
The line y=k is a horizontal asymptote of the graph of f if f(x)→k as x→∞ or as x→−∞.
The line y=mx+b is an oblique asymptote of the graph of f if the degree of N(x) is exactly one more than the degree of D(x).
A procedure for graphing rational functions is given on page 392.
Find the asymptotes off(x)=x2−x−6−x2+x+2.
Solution
The rational function f(x)=x2−x−6−x2+x+2=(x−3)(x+2)(x+1)(2−x) has domain (−∞,−1)∪(−1,2)∪(2,∞) because −1 and 2 are zeros of the denominator polynomial.
Because f is in lowest terms, the lines x=−1 and x=2 are vertical asymptotes of the graph of f.
The line y=1−1=−1 is a horizontal asymptote of the graph of f.
Find the asymptotes off(x)=x2+1x.
Solution
The line y=x is an oblique asymptote of the graph of the rational function f(x)=x2+1x=x+1x.
A procedure for graphing rational functions is given on page 392.
3.7 Variation
k is a nonzero constant called the constant of variation.
Variation
Equation
y varies directly with x.
y=kx
y varies with the nth power of x.
y=kxn
y varies inversely with x.
y=kx
y varies inversely with the nth power of x.
y=kxn
z varies jointly with the nth power of x and the mth power of y.
z=kxnym
z varies directly with the nth power of x and inversely with the mth power of y.
y=kxnym
Write an equation for each statement.
The distance s a body falls in t seconds is directly proportional to the square of the time t.
Solution
s=kt2
In a circuit with constant voltage, the current I varies inversely with the resistance R of the circuit.
Solution
I=kR
The volume V of a rectangular container of fixed length varies jointly with its depth d and width w.