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Zorluk: ZorPolynomial Factors and Graphs

The function ff is defined by f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15), where kk is a constant. In the xyxy-plane, the graph of y=f(x)y = f(x) is tangent to the xx-axis at the point (3,0)(3, 0). What is the value of kk?

Cevap: 3

Cevap

3
For the graph of a polynomial function to be tangent to the xx-axis at (3,0)(3, 0), the root x=3x = 3 must have an even multiplicity (at least 2). The function is defined as f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15). Since there is already one factor of (x3)(x - 3) explicitly defined, the remaining cubic factor g(x)=x3kx2+5x15g(x) = x^3 - kx^2 + 5x - 15 must also have a factor of (x3)(x - 3) to make the total multiplicity of the root x=3x = 3 at least 2. According to the Factor Theorem, if (x3)(x - 3) is a factor of g(x)g(x), then g(3)=0g(3) = 0. Substituting x=3x = 3 into g(x)g(x) gives 33k(3)2+5(3)15=03^3 - k(3)^2 + 5(3) - 15 = 0, which simplifies to 279k=027 - 9k = 0. Solving this equation for kk yields k=3k = 3.

Adım Adım Çözüm

1
Identify the relationship between graph tangency and factor multiplicity.
For the graph of a polynomial function to be tangent to the xx-axis at a point (c,0)(c, 0), the factor (xc)(x - c) must have an even multiplicity of at least 2 in the polynomial's factored form.
An odd multiplicity root causes the graph to cross the xx-axis, while an even multiplicity root causes the graph to touch the xx-axis and turn around (tangency).
2
Apply the multiplicity requirement to the given function.
Since f(x)=(x3)(x3kx2+5x15)f(x) = (x - 3)(x^3 - kx^2 + 5x - 15) already contains one factor of (x3)(x - 3), the cubic expression g(x)=x3kx2+5x15g(x) = x^3 - kx^2 + 5x - 15 must also contain (x3)(x - 3) as a factor to ensure the total multiplicity of the root x=3x = 3 is at least 2.
This guarantees that (x3)2(x - 3)^2 is a factor of f(x)f(x).
3
Apply the Factor Theorem to the cubic expression.
Since (x3)(x - 3) is a factor of g(x)g(x), then g(3)=0g(3) = 0.
The Factor Theorem states that a polynomial P(x)P(x) has a factor (xc)(x - c) if and only if P(c)=0P(c) = 0.
4
Solve for the constant kk by substituting x=3x = 3 into g(x)g(x).
33k(3)2+5(3)15=0    279k+1515=0    279k=0    9k=27    k=33^3 - k(3)^2 + 5(3) - 15 = 0 \implies 27 - 9k + 15 - 15 = 0 \implies 27 - 9k = 0 \implies 9k = 27 \implies k = 3.
Arithmetic simplification yields the value of the constant.

Anahtar Kavram

The relationship between polynomial factors, root multiplicities, and the behavior of the graph at xx-intercepts.
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