Let function f(x)=(x-a)(x-b). If f(10)=f(20)=0, f(30)=?
A. 50
B. 100
C. 150
D. 200
E. 250
OA is d
what approach can we use to get the correct answer?
Functions
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f(10) = 0 implies (10, 0) is a point on the graph of f(x)=(x-a)(x-b).Roland2rule wrote:Let function f(x)=(x-a)(x-b). If f(10)=f(20)=0, f(30)=?
A. 50
B. 100
C. 150
D. 200
E. 250
Implication:
f(x) = (x-10)(x-b), with the result that x=10 yields f(x)=0.
f(20) = 0 implies (20, 0) is a point on the graph of f(x)=(x-10)(x-b).
Implication:
f(x) = (x-10)(x-20), with the result that x=20 yields f(x)=0.
Substituting x=30 into f(x) = (x-10)(x-20), we get:
f(x) = (30-10)(30-20) = (20)(10) = 200.
The correct answer is D.
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Since f(10) = f(20) = 0, this means a = 10 and b = 20, or vice versa. In other words, f(x) = (x - 10)(x - 20). Therefore, f(30) = (30 - 10)(30 - 20) = (20)(10) = 200.BTGmoderatorRO wrote:Let function f(x)=(x-a)(x-b). If f(10)=f(20)=0, f(30)=?
A. 50
B. 100
C. 150
D. 200
E. 250
OA is d
what approach can we use to get the correct answer?
Answer: D
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