The function h(x) is quadratic and h(3) = h(–10) = 0. Which could represent h(x)?
A.h(x) = x2 – 13x – 30
B.h(x) = x2 – 7x – 30
C.h(x) = 2x2 + 26x – 60
D.h(x) = 2x2 + 14x – 60
Answers
Answer:
D. h(x) = 2x^2 + 14x - 60
Step-by-step explanation:
Given that h(x) is a quadratic.
Also, h(3) = h(-10) = 0
(A) h(x) = x^2 - 13x - 30
=> h(3) = 3^2 - 13(3) - 30
=> h(3) = 9 - 39 - 30
=> h(3) = -30 - 30
=> h(3) = -60
=> h(3) ≠ 0
(B) h(x) = x^2 - 7x - 30
=> h(3) = 3^2 - 7(3) - 30
=> h(3) = 9 - 21 - 30
=> h(3) = -12 - 30
=> h(3) = -42
=> h(3) ≠ 0
(C) h(x) = 2x^2 + 26x - 60
=> h(3) = 2(3^2) + 26(3) - 60
=> h(3) = 2(9) + 78 - 60
=> h(3) = 18 + 78 - 60
=> h(3) = 96 - 60
=> h(3) = 36
=> h(3) ≠ 0
(D) h(x) = 2x^2 + 14x - 60
=> h(3) = 2(3^2) + 14(3) - 60
=> h(3) = 2(9) + 42 - 60
=> h(3) = 18 + 42 - 60
=> h(3) = 60 - 60
=> h(3) = 0
And
=> h(-10) = 2(-10)^2 + 14(-10) - 60
=> h(-10) = 2(100) - 140 - 60
=> h(-10) = 200 - 200
=> h(-10) = 0
Clearly we have,
=> h(3) = h(-10) = 0
Hence, the correct option is (D) h(x) = 2x^2 + 14x - 60
Step-by-step explanation:
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Concept used here:-
Mathematics
Question:-
The function h(x) is quadratic and h(3) = h(–10) = 0. Which could represent h(x)?
A.h(x) = x2 – 13x – 30
B.h(x) = x2 – 7x – 30
C.h(x) = 2x2 + 26x – 60
D.h(x) = 2x2 + 14x – 60
Answer:-
Given,
h(3)=h(-10)
Such that we can understand that h(3) should replace in given four equations and then simplify answer should equals to -10.
Step.1 -
x²-13x-30
(3)²-13(3)-30
9-39-30
≠ 0
Step.2 -
x²-7x-30
(3)²-7(3)-30
9-21-30
≠ 0
Step.3 -
2x²+26x-60
2(3)²+26(3)-60
2(9)+78-60
≠ 0
Step.4 -
2x²+14x-60
2(3)²+14(3)-60
2(9)+14(3)-60
18+42-60
60-60
= 0