Question

If f(t)=4t²-3t+6, find 1.f(0) 2.f(4) 3.f(-5)​

Answer 1

Answer:

$f(t)=4t^{2}-3t+6$

1)

$f(0)=4(0)^{2}-3*0+6\\=0-0+6\\=0+6=6$

Answer is 6.

2)

$f(4)=4(4)^{2}-3*4+6\\=4*16-12+6\\=64-12+6\\=52+6=58$

Answer is 58.

3)

$f(-5)=4(-5)^{2}-3*(-5)+6\\=100-(-15)+6\\=100+15+6\\=115+6=121$

Answer is 121.

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Answer 2

Answer:

f(0)= 4×0^2 - 3×0+6

f(0)=0-0+6

f(0)=6

f(4)=4×4^2-3×4+6

f(4)=64 - 12+6

f(4)=58

f( -5) =4×-5^2-3×-5+6

f(-5)= -100+15+6

f(-5)= 79

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