The vertical _____ of a function secant are determined by the points that are not in the domain.
2 answers:
Answer:
The vertical <u>asymptote</u> of a function secant are determined by the points that are not in the domain.
Step-by-step explanation:
The domain of a function is the set of x values for which the function is defined.
Secant function is not defined at
It means we cannot include these points in the domain.
At these points, we must have a vertical line which do not touch the graph. These lines are called "Vertical asymptotes"
Vertical asymptotes are not included in the domain of the function.
Hence, the correct word should be "Asymptote"
The vertical <u>asymptote</u> of a function secant are determined by the points that are not in the domain.
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Answer:
-3 1/3
Step-by-step explanation:
The quadratic
... y = ax² +bx +c
has its extreme value at
... x = -b/(2a)
Since a = 3 is positive, we know the parabola opens upward and the extreme value is a minimum. (We also know that from the problem statement asking us to find the minimum value.) The value of x at the minimum is -(-4)/(2·3) = 2/3.
To find the minimum value, we need to evaluate the function for x=2/3.
The most straightforward way to do this is to substitue 2/3 for x.
... y = 3(2/3)² -4(2/3) -2 = 3(4/9) -8/3 -2
... y = (4 -8 -6)/3 = -10/3
... y = -3 1/3
_____
<em>Confirmation</em>
You can also use a graphing calculator to show you the minimum.
