Question

unit 3a: graphs of polynomial functions

1. ( f(x) = -x(x - 2)^2(x + 5)^5 )

zero:
multiplicity:
zero:
multiplicity:

1. for each graph, list the zeros and the multiplicity, the degree of the polynomial, and write a possible equation.

zeros and multiplicity:
degree
possible equation
5.
zero:
multiplicity:
degree:
zeros and multiplicity: ( x = -3 ) multiplicity, ( x = 0 ) multiplicity, ( x = 4 )
degree ( x = 4 )
possible equation ( f(x) = x(x + 3)(x - 4)^2 )

Explanation:

Step1: Analyze the graph (Problem 4)

The graph touches the x - axis at one point and crosses at another? Wait, no, looking at the first graph (Problem 4), the graph has a local maximum and minimum, and touches or crosses? Wait, the graph of a polynomial: when a zero has even multiplicity, the graph touches the x - axis (turns around), when odd, it crosses. Let's find the zeros. The graph intersects the x - axis at \(x=-1\) (crosses? No, wait, the graph: let's see the x - intercepts. Wait, the graph seems to have a zero at \(x = - 1\) (crossing? No, maybe \(x=0\)? Wait, no, the first graph (Problem 4) has a zero at \(x = - 1\) (multiplicity even? Wait, no, let's re - examine. Wait, the graph of the polynomial in Problem 4: let's count the number of turning points. The number of turning points of a polynomial of degree \(n\) is at most \(n - 1\). This graph has 2 turning points, so degree at least 3. Wait, but let's find zeros. The graph crosses the x - axis at \(x=-1\) (multiplicity odd) and touches at \(x = 0\) (multiplicity even)? Wait, no, maybe I misread. Wait, the first graph (Problem 4) has a zero at \(x = 0\) (touching, so multiplicity even, say 2) and a zero at \(x=-1\) (crossing, multiplicity odd, say 1). Then the degree is \(2 + 1=3\)? Wait, no, 2 (even) + 1 (odd)=3? Wait, no, the sum of multiplicities is the degree. If zero at \(x = 0\) with multiplicity 2 and zero at \(x=-1\) with multiplicity 1, degree is \(2 + 1=3\). Then the possible equation: \(f(x)=x^{2}(x + 1)\) (or with a leading coefficient, but since it's a possible equation, we can use leading coefficient 1). Wait, but let's check the y - intercept. When \(x = 0\), \(f(0)=0\), which matches.

Step1 (Problem 3): Given \(f(x)=-x(x - 2)^{2}(x + 5)^{5}\)

To find zeros: set \(f(x)=0\). So \(-x(x - 2)^{2}(x + 5)^{5}=0\). The zeros are \(x = 0\), \(x = 2\), \(x=-5\).

• For \(x = 0\): The exponent of \((x - 0)\) is 1, so multiplicity 1 (odd, so the graph crosses the x - axis at \(x = 0\)).
• For \(x = 2\): The exponent of \((x - 2)\) is 2, so multiplicity 2 (even, so the graph touches the x - axis at \(x = 2\)).
• For \(x=-5\): The exponent of \((x + 5)=(x-(-5))\) is 5, so multiplicity 5 (odd, so the graph crosses the x - axis at \(x=-5\)).

The degree of the polynomial is the sum of the exponents: \(1+2 + 5=8\).

Step1 (Problem 5): The second graph (Problem 5)

Looking at the graph, the zeros: let's see the x - intercepts. The graph crosses the x - axis at \(x=-3\), \(x = 0\), and touches at \(x = 4\) (since it turns around at \(x = 4\)). So:

• Zero at \(x=-3\): multiplicity 1 (odd, crosses)
• Zero at \(x = 0\): multiplicity 1 (odd, crosses)
• Zero at \(x = 4\): multiplicity 2 (even, touches)

Degree: \(1+1 + 2=4\)

Possible equation: \(f(x)=x(x + 3)(x - 4)^{2}\) (leading coefficient 1, y - intercept at 0 when \(x = 0\), which matches the graph touching or crossing at those points).

Answer:

Problem 3:

• Zero: \(x = 0\), Multiplicity: 1; Zero: \(x = 2\), Multiplicity: 2; Zero: \(x=-5\), Multiplicity: 5; Degree: 8.

Problem 4:

• Zeros and multiplicity: \(x = 0\) (multiplicity 2), \(x=-1\) (multiplicity 1); Degree: 3; Possible Equation: \(f(x)=x^{2}(x + 1)\) (or with a non - zero leading coefficient, e.g., \(f(x)=2x^{2}(x + 1)\) etc.)

Problem 5:

• Zeros and multiplicity: \(x=-3\) (multiplicity 1), \(x = 0\) (multiplicity 1), \(x = 4\) (multiplicity 2); Degree: 4; Possible Equation: \(f(x)=x(x + 3)(x - 4)^{2}\) (or with a non - zero leading coefficient)