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Solve for j(h(x)):
h(x) = x² + 4
j(x) = 4x - 1
Substitute in h(x) into "x":
j(h(x)) = 4(x² + 4) - 1
j(h(x)) = 4x² + 16 - 1
j(h(x)) = 4x² + 15
Solve at x = 2:
4(2)² + 15 = 4(4) + 15 = 31.
Subtract 6.1 and 51.24
1 answer:
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ANSWER
Frequency=2
Midline =3
Amplitude =5
Period =π
EXPLANATION
We have:
The given function is of the form;
where
a=|5| =5 is the amplitude
and b=2 is the frequency.
and 2π/b=2π/2=π
The range of the given function is
8≤y≤-2
The midline is