Question

1. sketch a graph ( f(x) ) such that ( f(x) = 0 ) has the indicated number and type of solutions.

a. 5 real
b. 3 real, 2 imaginary
c. 4 imaginary

Explanation:

Part a: 5 real solutions

Step 1: Recall Polynomial Degree

A polynomial with \( n \) real roots (solutions to \( f(x) = 0 \)) must have degree at least \( n \), and real roots correspond to \( x \)-intercepts. For 5 real solutions, use a 5th-degree polynomial (odd degree, end - behaviors: \( x\to\infty, f(x)\to\infty \); \( x\to-\infty, f(x)\to-\infty \) or vice - versa).

Step 2: Construct the Polynomial

Let \( f(x)=(x - a)(x - b)(x - c)(x - d)(x - e) \), where \( a,b,c,d,e \) are distinct real numbers (e.g., \( a=-2,b = - 1,c = 0,d = 1,e = 2 \)).

Step 3: Sketch the Graph

• The graph will cross the \( x \)-axis at \( x=-2,-1,0,1,2 \).
• For a 5th - degree polynomial with positive leading coefficient, as \( x\to\infty \), \( f(x)\to\infty \); as \( x\to-\infty \), \( f(x)\to-\infty \). The graph will have 4 turning points (since degree \( n \) has at most \( n - 1 \) turning points) between the \( x \)-intercepts, creating a “wavy” shape crossing the \( x \)-axis 5 times.

Part b: 3 real, 2 imaginary solutions

Step 1: Recall Polynomial Degree and Roots

Imaginary roots come in conjugate pairs (for polynomials with real coefficients). So, 2 imaginary roots mean a quadratic factor with no real roots (\( ax^{2}+bx + c \), \( b^{2}-4ac<0 \)). The total degree is \( 3 + 2=5 \) (since 3 real roots + 2 imaginary roots (1 quadratic factor) gives a 5th - degree polynomial).

Step 2: Construct the Polynomial

Let \( f(x)=(x - a)(x - b)(x - c)(x^{2}+dx + e) \), where \( a,b,c \) are real (e.g., \( a=-1,b = 0,c = 1 \)) and \( x^{2}+dx + e \) has \( d^{2}-4e<0 \) (e.g., \( x^{2}+x + 1 \), since \( 1^{2}-4(1)=-3<0 \)).

Step 3: Sketch the Graph

• The graph crosses the \( x \)-axis at \( x=-1,0,1 \) (3 real roots).
• The quadratic factor \( x^{2}+x + 1 \) has no real roots, so the graph does not cross the \( x \)-axis for the part corresponding to this factor. The 5th - degree polynomial (with positive leading coefficient) has end - behaviors \( x\to\infty, f(x)\to\infty \); \( x\to-\infty, f(x)\to-\infty \), and 2 turning points between the real roots (since degree 5, at most 4 turning points, and 3 real roots + 1 quadratic factor will have a combination of turning points from the linear and quadratic parts).

Part c: 4 imaginary solutions

Step 1: Recall Polynomial Degree and Roots

Imaginary roots come in conjugate pairs, so 4 imaginary roots mean two quadratic factors with no real roots (\( (x^{2}+a)(x^{2}+b) \), \( a,b>0 \) to ensure \( x^{2}+a = 0 \) and \( x^{2}+b = 0 \) have no real roots). The total degree is \( 2 + 2=4 \) (even degree).

Step 2: Construct the Polynomial

Let \( f(x)=(x^{2}+1)(x^{2}+4) \) (both \( x^{2}+1 = 0 \) (roots \( \pm i \)) and \( x^{2}+4 = 0 \) (roots \( \pm 2i \)) have no real roots).

Step 3: Sketch the Graph

• For a 4th - degree polynomial with positive leading coefficient, as \( x\to\pm\infty \), \( f(x)\to\infty \).
• Since there are no real roots, the graph never crosses the \( x \)-axis. It will have a “U - shaped” or “W - shaped” form (with 3 turning points at most, for degree 4) above or below the \( x \)-axis. For \( f(x)=(x^{2}+1)(x^{2}+4)=x^{4}+5x^{2}+4 \), the minimum value occurs at \( x = 0 \), \( f(0)=4>0 \), so the graph is always above the \( x \)-axis, with a “W” shape (two minima and one maximum) or similar, never crossing the \( x \)-axis.

Answer:

s (Graph Descriptions):
a. A 5th - degree polynomial (e.g., \( f(x)=(x + 2)(x + 1)x(x - 1)(x - 2) \)) crossing the \( x \)-axis 5 times.
b. A 5th - degree polynomial (e.g., \( f(x)=(x + 1)x(x - 1)(x^{2}+x + 1) \)) crossing the \( x \)-axis 3 times and having a non - real quadratic factor.
c. A 4th - degree polynomial (e.g., \( f(x)=(x^{2}+1)(x^{2}+4) \)) never crossing the \( x \)-axis.