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Zorluk: OrtaPolynomial Factors and Graphs

A polynomial function ff has the form f(x)=a(x2)(x+3)(x5)f(x) = a(x - 2)(x + 3)(x - 5), where aa is a constant. In the xyxy-plane, the graph of y=f(x)y = f(x) has a yy-intercept of (0,60)(0, 60). What is the value of f(1)f(1)?

Cevap: 32

Cevap

32
The yy-intercept of the graph is given as (0,60)(0, 60), which means that when x=0x = 0, f(0)=60f(0) = 60. Substituting x=0x = 0 into the function gives f(0)=a(02)(0+3)(05)=30af(0) = a(0 - 2)(0 + 3)(0 - 5) = 30a. Equating this to 60 gives 30a=6030a = 60, so a=2a = 2. Therefore, the function is f(x)=2(x2)(x+3)(x5)f(x) = 2(x - 2)(x + 3)(x - 5). To find the value of f(1)f(1), substitute x=1x = 1 into this expression: f(1)=2(12)(1+3)(15)=2(1)(4)(4)=32f(1) = 2(1 - 2)(1 + 3)(1 - 5) = 2(-1)(4)(-4) = 32.

Adım Adım Çözüm

1
Identify the relation between the yy-intercept and the function's value.
f(0)=60f(0) = 60
The yy-intercept of a graph y=f(x)y = f(x) is the point where x=0x = 0.
2
Substitute x=0x = 0 into the definition of f(x)f(x) and set it equal to 60.
a(02)(0+3)(05)=60    30a=60a(0 - 2)(0 + 3)(0 - 5) = 60 \implies 30a = 60
This allows us to solve for the unknown constant coefficient aa.
3
Solve the linear equation for aa.
a=2a = 2
Dividing both sides of the equation by 30 isolates aa.
4
Evaluate the complete function f(x)=2(x2)(x+3)(x5)f(x) = 2(x - 2)(x + 3)(x - 5) at x=1x = 1.
f(1)=2(12)(1+3)(15)=2(1)(4)(4)=32f(1) = 2(1 - 2)(1 + 3)(1 - 5) = 2(-1)(4)(-4) = 32
This yields the requested value of f(1)f(1).

Anahtar Kavram

Using the factors and a known point (such as the y-intercept) of a polynomial function to determine its algebraic expression and evaluate it.
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