Find the standard form of the quadratic equation: f(x)=(2x-1)(x+4)
1 answer:
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Answer:
x >= -7 ................(1a)
x >= 3 ...............(2a1)
Step-by-step explanation:
f(x) = .............(0)
find the domain.
To find the (real) domain, we need to ensure that each term remains a real number.
which means the following conditions must be met
x+7 >= 0 .....................(1)
AND
x^2+2x-15 >= 0 ..........(2)
To satisfy (1), x >= -7 .....................(1a) by transposition of (1)
To satisfy (2), we need first to find the roots of (2)
factor (2)
(x+5)(x-3) >= 0
This implis
(x+5) >= 0 AND (x-3) >= 0.....................(2a)
OR
(x+5) <= 0 AND (x-3) <= 0 ...................(2b)
(2a) is satisfied with x >= 3 ...............(2a1)
(2b) is satisfied with x <= -5 ................(2b1)
Combine the conditions (1a), (2a1), and (2b1),
x >= -7 ................(1a)
AND
(
x >= 3 ...............(2a1)
OR
x <= -5 ................(2b1)
)
which expands to
(1a) and (2a1) OR (1a) and (2b1)
( x >= -7 and x >= 3 ) OR ( x >= -7 and x <= -5 )
Simplifying, we have
x >= 3 OR ( -7 <= x <= -5 )
Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.
Observe the given figure.
Here, angle T and R are the inscribed angles.
And sum of inscribed angles is half of measure of intercepted arcs.
Since, the measure of intercepted arcs is 360 degrees.
Therefore,
Hence, the sum of the two angles is 180 degrees.
Therefore, these are the supplementary angles.
Therefore, angle T and angle R are supplementary angles.
Hence, proved.