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Zorluk: OrtaPolynomial Factors and Graphs

A polynomial f(x)f(x) is defined by f(x)=2x3+hx27x6f(x) = 2x^3 + hx^2 - 7x - 6, where hh is a constant. If the graph of y=f(x)y = f(x) in the xyxy-plane passes through the point (2,0)(2, 0), what is the value of hh?

Cevap: 1

Cevap

The value of the constant hh is 1.
Since the graph of y=f(x)y = f(x) passes through the point (2,0)(2, 0), the value of the function at x=2x = 2 must be 00. Substituting x=2x = 2 into the equation yields 2(2)3+h(2)27(2)6=02(2)^3 + h(2)^2 - 7(2) - 6 = 0. Simplifying the terms gives 2(8)+4h146=02(8) + 4h - 14 - 6 = 0, which simplifies to 16+4h20=016 + 4h - 20 = 0, or 4h4=04h - 4 = 0. Solving for hh yields h=1h = 1.

Adım Adım Çözüm

1
Apply the point condition to the polynomial function.
f(2)=0f(2) = 0
Since the point (2,0)(2, 0) lies on the graph of y=f(x)y = f(x), substituting x=2x = 2 must yield y=0y = 0.
2
Substitute x=2x = 2 into the expression for f(x)f(x).
f(2)=2(2)3+h(2)27(2)6f(2) = 2(2)^3 + h(2)^2 - 7(2) - 6
We replace each occurrence of xx with 22 to evaluate the function value.
3
Simplify the algebraic expression.
16+4h20=0    4h4=016 + 4h - 20 = 0 \implies 4h - 4 = 0
Evaluating the exponents and multiplications gives 2(8)+4h146=16+4h20=4h42(8) + 4h - 14 - 6 = 16 + 4h - 20 = 4h - 4.
4
Solve the linear equation for hh.
h=1h = 1
Adding 4 to both sides gives 4h=44h = 4, and dividing by 4 gives h=1h = 1.

Anahtar Kavram

Connecting graphical x-intercepts of a polynomial to its algebraic roots and evaluation
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