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3. sketch and show all the possible scenarios for the number of solutio…

Question

  1. sketch and show all the possible scenarios for the number of solutions. first is done for you.

a. positive cubic:

b. negative quartic:

c. positive quadratic:

d. negative quintic:

2024-11-15
page 2 of 20

Explanation:

Response
Part b: Negative Quartic (Degree 4, Leading Coefficient Negative)

A quartic function has the form \( y = ax^4 + bx^3 + cx^2 + dx + e \), where \( a
eq 0 \). For a negative quartic, \( a < 0 \), so as \( x \to \pm\infty \), \( y \to -\infty \) (since the leading term \( ax^4 \) dominates, and \( x^4 \) is always non - negative, so negative times non - negative is non - positive, and as \( |x| \) becomes large, \( y \) becomes very negative).

1 solution (1 sol'n)

The graph of the negative quartic is always below the x - axis (or just touches the x - axis at one point). So, we draw a curve that starts from \( -\infty \) as \( x\to-\infty \), comes up, touches the x - axis at one point, and then goes back to \( -\infty \) as \( x\to\infty \).

2 solutions (2 sol'ns)

The graph of the negative quartic crosses the x - axis at two distinct points. It starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis at one point, reaches a local maximum (above the x - axis), then falls, crosses the x - axis again, and then goes back to \( -\infty \) as \( x\to\infty \).

3 solutions (3 sol'ns)

A quartic function is a degree 4 polynomial. By the Fundamental Theorem of Algebra, a degree \( n \) polynomial has at most \( n \) real roots. But for a negative quartic, we can have a graph that starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis, reaches a local maximum, falls, crosses the x - axis again, reaches a local minimum (above the x - axis), rises again, crosses the x - axis a third time, and then falls back to \( -\infty \) as \( x\to\infty \). Wait, actually, a quartic can have 0, 2, or 4 real roots? Wait, no, the number of real roots of a quartic (degree 4) can be 0, 2, or 4? Wait, no, the possible number of real roots for a quartic: since it's a continuous function, and the end - behavior is \( y\to-\infty \) as \( x\to\pm\infty \). The number of times it can cross the x - axis: 0 (if the graph is always below the x - axis), 2 (if it has a "hill" above the x - axis and crosses twice), or 4 (if it has a "hill" and a "valley" above the x - axis, so it crosses the x - axis four times). Wait, maybe the original problem has a typo, but following the pattern of the given problem (where for cubic (degree 3) we have 1, 2, 3 solutions), for quartic (degree 4), we will sketch as per the problem's expectation. Let's assume that the problem considers the number of real roots (solutions) as per the given structure. So, for 3 sol'ns, we draw a curve that crosses the x - axis three times (even though mathematically a quartic can have at most 4 real roots, but we follow the problem's pattern). The curve starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis, reaches a local maximum, falls, crosses the x - axis, reaches a local minimum (above the x - axis), rises, crosses the x - axis, reaches a local maximum, and then falls back to \( -\infty \) as \( x\to\infty \).

Part c: Positive Quadratic (Degree 2, Leading Coefficient Positive)

A positive quadratic function has the form \( y = ax^2+bx + c \), where \( a>0 \). The graph of a positive quadratic is a parabola opening upwards.

1 solution (1 sol'n)

The parabola touches the x - axis at exactly one point (the discriminant \( D=b^2 - 4ac = 0 \)). So, the vertex of the parabola lies on the x - axis. The equation of the parabola can be written as \( y=a(x - h)^2 \), where \( (h,0) \) is the vertex.

2 solutions (2 sol'ns)

The parabola crosses the x - axis at two distinct points (the discriminant \( D = b^2-4ac>0 \)).…

Answer:

Part b: Negative Quartic (Degree 4, Leading Coefficient Negative)

A quartic function has the form \( y = ax^4 + bx^3 + cx^2 + dx + e \), where \( a
eq 0 \). For a negative quartic, \( a < 0 \), so as \( x \to \pm\infty \), \( y \to -\infty \) (since the leading term \( ax^4 \) dominates, and \( x^4 \) is always non - negative, so negative times non - negative is non - positive, and as \( |x| \) becomes large, \( y \) becomes very negative).

1 solution (1 sol'n)

The graph of the negative quartic is always below the x - axis (or just touches the x - axis at one point). So, we draw a curve that starts from \( -\infty \) as \( x\to-\infty \), comes up, touches the x - axis at one point, and then goes back to \( -\infty \) as \( x\to\infty \).

2 solutions (2 sol'ns)

The graph of the negative quartic crosses the x - axis at two distinct points. It starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis at one point, reaches a local maximum (above the x - axis), then falls, crosses the x - axis again, and then goes back to \( -\infty \) as \( x\to\infty \).

3 solutions (3 sol'ns)

A quartic function is a degree 4 polynomial. By the Fundamental Theorem of Algebra, a degree \( n \) polynomial has at most \( n \) real roots. But for a negative quartic, we can have a graph that starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis, reaches a local maximum, falls, crosses the x - axis again, reaches a local minimum (above the x - axis), rises again, crosses the x - axis a third time, and then falls back to \( -\infty \) as \( x\to\infty \). Wait, actually, a quartic can have 0, 2, or 4 real roots? Wait, no, the number of real roots of a quartic (degree 4) can be 0, 2, or 4? Wait, no, the possible number of real roots for a quartic: since it's a continuous function, and the end - behavior is \( y\to-\infty \) as \( x\to\pm\infty \). The number of times it can cross the x - axis: 0 (if the graph is always below the x - axis), 2 (if it has a "hill" above the x - axis and crosses twice), or 4 (if it has a "hill" and a "valley" above the x - axis, so it crosses the x - axis four times). Wait, maybe the original problem has a typo, but following the pattern of the given problem (where for cubic (degree 3) we have 1, 2, 3 solutions), for quartic (degree 4), we will sketch as per the problem's expectation. Let's assume that the problem considers the number of real roots (solutions) as per the given structure. So, for 3 sol'ns, we draw a curve that crosses the x - axis three times (even though mathematically a quartic can have at most 4 real roots, but we follow the problem's pattern). The curve starts from \( -\infty \) as \( x\to-\infty \), rises, crosses the x - axis, reaches a local maximum, falls, crosses the x - axis, reaches a local minimum (above the x - axis), rises, crosses the x - axis, reaches a local maximum, and then falls back to \( -\infty \) as \( x\to\infty \).

Part c: Positive Quadratic (Degree 2, Leading Coefficient Positive)

A positive quadratic function has the form \( y = ax^2+bx + c \), where \( a>0 \). The graph of a positive quadratic is a parabola opening upwards.

1 solution (1 sol'n)

The parabola touches the x - axis at exactly one point (the discriminant \( D=b^2 - 4ac = 0 \)). So, the vertex of the parabola lies on the x - axis. The equation of the parabola can be written as \( y=a(x - h)^2 \), where \( (h,0) \) is the vertex.

2 solutions (2 sol'ns)

The parabola crosses the x - axis at two distinct points (the discriminant \( D = b^2-4ac>0 \)). The parabola opens upwards, starts from \( +\infty \) as \( x\to\pm\infty \), and crosses the x - axis at two points \( x_1 \) and \( x_2 \) (where \( x_1

3 solutions (3 sol'ns)

A quadratic function (degree 2) can have at most 2 real roots (by the Fundamental Theorem of Algebra, a degree \( n \) polynomial has at most \( n \) real roots). So, a positive quadratic cannot have 3 real roots. This is a mistake in the problem, but if we follow the problem's pattern (maybe a mis - labeling, and it should be a cubic or another function), but if we assume that we have to draw it, we can note that it's impossible for a quadratic to have 3 real roots. However, if we ignore the mathematical accuracy for the sake of following the problem's structure, we can try to draw a curve that looks like a parabola but crosses the x - axis three times, but this is not a quadratic curve.

Part d: Negative Quintic (Degree 5, Leading Coefficient Negative)

A quintic function has the form \( y=ax^5+bx^4+cx^3+dx^2+ex + f \), where \( a
eq0 \). For a negative quintic, \( a < 0 \), so as \( x\to\infty \), \( y\to-\infty \) (since \( x^5\to\infty \) as \( x\to\infty \), and \( a<0 \) so \( ax^5\to-\infty \)) and as \( x\to-\infty \), \( y\to\infty \) (since \( x^5\to-\infty \) as \( x\to-\infty \), and \( a < 0 \) so \( ax^5\to\infty \)).

1 solution (1 sol'n)

The graph of the negative quintic crosses the x - axis at one point. It starts from \( +\infty \) as \( x\to-\infty \), decreases, crosses the x - axis at one point, and then continues to decrease to \( -\infty \) as \( x\to\infty \).

2 solutions (2 sol'ns)

The graph of the negative quintic crosses the x - axis at two points. It starts from \( +\infty \) as \( x\to-\infty \), decreases, crosses the x - axis, reaches a local minimum (below the x - axis), increases, crosses the x - axis again, and then decreases to \( -\infty \) as \( x\to\infty \).

3 solutions (3 sol'ns)

The graph of the negative quintic crosses the x - axis at three points. It starts from \( +\infty \) as \( x\to-\infty \), decreases, crosses the x - axis, reaches a local minimum, increases, crosses the x - axis, reaches a local maximum (above the x - axis), decreases, crosses the x - axis, and then decreases to \( -\infty \) as \( x\to\infty \). A quintic (degree 5) polynomial can have 1, 3, or 5 real roots (since it's an odd - degree polynomial, it must have at least 1 real root, and the number of real roots is odd, so 1, 3, or 5).

To actually sketch the graphs:

Negative Quartic (b)
  • 1 sol'n: Draw a curve that starts from the bottom left ( \( x\to-\infty,y\to-\infty \) ), rises, touches the x - axis at one point, and then falls back to the bottom right ( \( x\to\infty,y\to-\infty \) ).
  • 2 sol'ns: Draw a curve that starts from the bottom left, rises, crosses the x - axis, reaches a peak (above the x - axis), falls, crosses the x - axis, and then falls back to the bottom right.
  • 3 sol'ns: Draw a curve that starts from the bottom left, rises, crosses the x - axis, reaches a peak, falls, crosses the x - axis, reaches a valley (above the x - axis), rises, crosses the x - axis, and then falls back to the bottom right.
Positive Quadratic (c)
  • 1 sol'n: Draw a parabola opening upwards with its vertex on the x - axis.
  • 2 sol'ns: Draw a parabola opening upwards that crosses the x - axis at two distinct points.
  • 3 sol'ns: (Mathematically impossible for a quadratic, but if we follow the problem's structure) Draw a curve that looks like a parabola but crosses the x - axis three times (this is a non - quadratic curve, maybe the problem meant a cubic, but we proceed as per instruction).
Negative Quintic (d)
  • 1 sol'n: Draw a curve that starts from the top left ( \( x\to-\infty,y\to\infty \) ), decreases, crosses the x - axis at one point, and then decreases to the bottom right ( \( x\to\infty,y\to-\infty \) ).
  • 2 sol'ns: Draw a curve that starts from the top left, decreases, crosses the x - axis, reaches a local minimum (below the x - axis), increases, crosses the x - axis, and then decreases to the bottom right.
  • 3 sol'ns: Draw a curve that starts from the top left, decreases, crosses the x - axis, reaches a local minimum, increases, crosses the x - axis, reaches a local maximum (above the x - axis), decreases, crosses the x - axis, and then decreases to the bottom right.

Since the problem asks to sketch and show all possible scenarios, the key is to use the end - behavior of the polynomials (based on their degree and leading coefficient) and the fact that polynomials are continuous to determine the number of times they can cross the x - axis (number of solutions).

For the answer, since it's a sketching problem, we can describe the general shape of each graph as above. But if we have to summarize the number of solutions:

  • Negative Quartic: Possible number of real solutions (based on the problem's pattern) 1, 2, 3 (even though mathematically a quartic can have 0, 2, 4 real roots, we follow the problem's given structure).
  • Positive Quadratic: A quadratic can have 0, 1, or 2 real roots. The problem's 3 - solution case is mathematically incorrect for a quadratic.
  • Negative Quintic: A quintic (odd degree) can have 1, 3, or 5 real roots. The problem's 2 - solution case is mathematically incorrect (since an odd - degree polynomial must have an odd number of real roots), but we follow the problem's pattern.

If we consider the problem's requirement to sketch the graphs:

For part (b) Negative Quartic:

  • 1 sol'n: Curve touching x - axis at 1 point, opening down (since leading coefficient negative) with end - behavior \( y\to-\infty \) as \( x\to\pm\infty \).
  • 2 sol'ns: Curve crossing x - axis at 2 points, with a peak above x - axis.
  • 3 sol'ns: Curve crossing x - axis at 3 points, with a peak and a valley above x - axis.

For part (c) Positive Quadratic:

  • 1 sol'n: Parabola opening up, vertex on x - axis.
  • 2 sol'ns: Parabola opening up, crossing x - axis at 2 points.
  • 3 sol'ns: (Incorrect for quadratic, but as per problem) A curve like a parabola but crossing x - axis 3 times (not a quadratic).

For part (d) Negative Quintic:

  • 1 sol'n: Curve starting from top left, crossing x - axis once, going to bottom right.
  • 2 sol'ns: (Incorrect for odd - degree, but as per problem) Curve starting from top left, crossing x - axis twice (but mathematically should be odd number).
  • 3 sol'ns: Curve starting from top left, crossing x - axis three times, with a peak above x - axis.

The main idea is to use the degree of the polynomial (to know the maximum number of roots) and the leading coefficient (to know the end - behavior) to sketch the possible graphs.