Question

c. $f(x) = (x^2 + 1)(x - 2)(x - 3)$

degree: _______________

leading coefficient (+ or -): ______

zeros: _______________

multiplicities: _______________

y – intercept: _______________

4.
which of the following is the polynomial equation of
degree 4 that sketches the graph of $f$ shown to the
right?

a. $f(x) = \frac{1}{2}x^3(2x - 5)$

b. $f(x) = \frac{1}{2}x^3(2x + 5)$

c. $f(x) = x^3(2x - 5)$

d. $f(x) = x^3(2x + 5)$

5.
which graph has the following characteristics?

• three real zeros
• as $x \to -\infty, f(x) \to -\infty$
• as $x \to \infty, f(x) \to \infty$

a.
graph a: a curve with arrows, crossing x-axis, with end behaviors: left arrow up, right arrow down

b.
graph b: a curve with arrows, crossing x-axis, with end behaviors: both arrows up

c.
graph c: a curve with arrows, crossing x-axis, with end behaviors: left arrow down, right arrow up

d.
graph d: a curve with arrows, crossing x-axis, with end behaviors: both arrows down

Explanation:

Problem c:

Step1: Find the degree

To find the degree of the polynomial \( f(x)=(x^{2}+1)(x - 2)(x - 3) \), we first expand the product. The degree of a polynomial is the highest power of \( x \) with a non - zero coefficient. When we multiply the factors, the highest power of \( x \) comes from multiplying the highest powers of \( x \) in each factor. The factor \( x^{2}+1 \) has a highest power of \( x^{2} \), and the factors \( (x - 2) \) and \( (x - 3) \) have a highest power of \( x \) each. So, when we multiply \( x^{2}\times x\times x=x^{4} \). So the degree of the polynomial is 4.

Step2: Find the leading coefficient

The leading coefficient is the coefficient of the term with the highest degree. When we expand \( (x^{2}+1)(x - 2)(x - 3) \), the leading term comes from \( x^{2}\times x\times x=x^{4} \) (since the coefficients of \( x^{2} \) in \( x^{2}+1 \), \( x \) in \( x - 2 \) and \( x \) in \( x - 3 \) are all 1). So the leading coefficient is positive (+).

Step3: Find the zeros

To find the zeros, we set \( f(x) = 0 \), i.e., \( (x^{2}+1)(x - 2)(x - 3)=0 \). Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So we set each factor equal to zero:

• For \( x^{2}+1=0 \), we have \( x^{2}=- 1 \), and the solutions are \( x = i \) and \( x=-i \) (complex zeros).
• For \( x - 2=0 \), we have \( x = 2 \).
• For \( x - 3=0 \), we have \( x = 3 \).

The real zeros are \( x = 2 \) and \( x = 3 \), and the complex zeros are \( x = i \) and \( x=-i \). But usually, when we talk about zeros in the context of real - valued polynomials (for graphing etc.), we consider real zeros first. So the real zeros are 2 and 3, and complex zeros are \( i,-i \).

Step4: Find the multiplicities

The multiplicity of a zero is the number of times the corresponding factor appears in the factored form of the polynomial. For the factor \( (x - 2) \), it appears once, so the multiplicity of \( x = 2 \) is 1. For the factor \( (x - 3) \), it appears once, so the multiplicity of \( x = 3 \) is 1. For the factor \( (x^{2}+1)=(x - i)(x + i) \), each of the factors \( (x - i) \) and \( (x + i) \) appears once, so the multiplicity of \( x = i \) and \( x=-i \) is 1.

Step5: Find the y - intercept

To find the y - intercept, we set \( x = 0 \) in the function \( f(x) \). So \( f(0)=(0^{2}+1)(0 - 2)(0 - 3)=(1)\times(-2)\times(-3)=6 \). So the y - intercept is 6.

1. First, check the degree: The polynomial should be of degree 4. Let's expand each option:

• Option A: \( f(x)=\frac{1}{2}x^{3}(2x - 5)=x^{4}-\frac{5}{2}x^{3} \), degree 4.
• Option B: \( f(x)=\frac{1}{2}x^{3}(2x + 5)=x^{4}+\frac{5}{2}x^{3} \), degree 4.
• Option C: \( f(x)=x^{3}(2x - 5)=2x^{4}-5x^{3} \), degree 4.
• Option D: \( f(x)=x^{3}(2x + 5)=2x^{4}+5x^{3} \), degree 4.

1. Then, check the zeros: The graph has a zero at \( x = 0 \) (with multiplicity 3, since the graph touches and turns around at \( x = 0 \) or crosses with a flat tangent? Wait, the graph passes through the origin and has a zero at \( x=\frac{5}{2}\) or \( x =-\frac{5}{2}\)? Wait, let's find the zeros of each function:

• For option A: \( f(x)=\frac{1}{2}x^{3}(2x - 5)=0 \) when \( x^{3}=0 \) (i.e., \( x = 0 \) with multiplicity 3) or \( 2x-5 = 0\Rightarrow x=\frac{5}{2}\).
• For option B: \( f(x)=\frac{1}{2}x^{3}(2x + 5)=0 \) when \( x^{3}=0 \) (i.e., \( x = 0 \) with multiplicity 3) or \( 2x + 5=0\Rightarrow x=-\frac{5}{2}\).
• For option C: \( f(x)=x^{3}(2x - 5)=0 \) when \( x^{3}=0 \) (i.e., \( x = 0 \) with multiplicity 3) or \( 2x-5 = 0\Rightarrow x=\frac{5}{2}\).
• For option D: \( f(x)=x^{3}(2x + 5)=0 \) when \( x^{3}=0 \) (i.e., \( x = 0 \) with multiplicity 3) or \( 2x + 5=0\Rightarrow x=-\frac{5}{2}\).

1. Now, check the point \( (2,-4) \): Let's substitute \( x = 2 \) into each function:

• Option A: \( f(2)=\frac{1}{2}(2)^{3}(2\times2 - 5)=\frac{1}{2}\times8\times(4 - 5)=4\times(-1)=-4 \).
• Option B: \( f(2)=\frac{1}{2}(2)^{3}(2\times2 + 5)=\frac{1}{2}\times8\times(4 + 5)=4\times9 = 36

eq-4 \).

• Option C: \( f(2)=2^{3}(2\times2 - 5)=8\times(4 - 5)=8\times(-1)=-8

eq-4 \).

• Option D: \( f(2)=2^{3}(2\times2 + 5)=8\times(4 + 5)=8\times9 = 72

eq-4 \).

So the correct function is \( f(x)=\frac{1}{2}x^{3}(2x - 5) \).

1. Analyze the end - behavior: We know that for a polynomial function \( y = f(x)=a_{n}x^{n}+a_{n - 1}x^{n - 1}+\cdots+a_{0} \), the end - behavior is determined by the leading term \( a_{n}x^{n} \).

• The given end - behavior is as \( x

ightarrow-\infty,f(x)
ightarrow-\infty \) and as \( x
ightarrow\infty,f(x)
ightarrow\infty \). This means that the degree of the polynomial is odd and the leading coefficient is positive. Because for an odd - degree polynomial \( y = a_{n}x^{n} \), if \( a_{n}>0 \), then as \( x
ightarrow-\infty,y
ightarrow-\infty \) and as \( x
ightarrow\infty,y
ightarrow\infty \); if \( a_{n}<0 \), then as \( x
ightarrow-\infty,y
ightarrow\infty \) and as \( x
ightarrow\infty,y
ightarrow-\infty \).

• Also, the polynomial has three real zeros.

1. Analyze each graph:

• Option A: The end - behavior: As \( x

ightarrow-\infty \), the graph goes up ( \( f(x)
ightarrow\infty \) ) and as \( x
ightarrow\infty \), the graph goes down ( \( f(x)
ightarrow-\infty \) ). So the leading coefficient is negative and the degree is odd. This does not match the given end - behavior.

• Option B: The end - behavior: As \( x

ightarrow-\infty \), the graph goes up ( \( f(x)
ightarrow\infty \) ) and as \( x
ightarrow\infty \), the graph goes up ( \( f(x)
ightarrow\infty \) ). So the degree is even (since for even - degree polynomials, the ends go in the same direction). This does not match the requirement of odd degree.

• Option C: The end - behavior: As \( x

ightarrow-\infty \), the graph goes down ( \( f(x)
ightarrow-\infty \) ) and as \( x
ightarrow\infty \), the graph goes up ( \( f(x)
ightarrow\infty \) ). The degree is odd (since the ends go in opposite directions) and the leading coefficient is positive (because as \( x
ightarrow\infty,f(x)
ightarrow\infty \)). Also, it has three real zeros (it crosses the x - axis three times).

• Option D: The end - behavior: As \( x

ightarrow-\infty \), the graph goes down ( \( f(x)
ightarrow-\infty \) ) and as \( x
ightarrow\infty \), the graph goes down ( \( f(x)
ightarrow-\infty \) ). So the degree is even (ends go in the same direction) and the leading coefficient is negative. This does not match the given end - behavior.

Answer:

Degree: \( \boldsymbol{4} \)
Leading Coefficient (+ or -): \( \boldsymbol{+} \)
Zeros: \( \boldsymbol{x = 2,x = 3,x = i,x=-i} \) (real zeros: 2, 3; complex zeros: \( i,-i \))
Multiplicities: \( \boldsymbol{1} \) (for \( x = 2 \), \( x = 3 \), \( x = i \), \( x=-i \))
Y - Intercept: \( \boldsymbol{6} \)

Problem 4: