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2 answers:
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Graph attached.
To graph by hand find the values at 1 and 3 by plugging it into the original function and drawing a line between the two.
2(1)-2 = 0 Point (1,0)
2(3)-2 = 4 Point (3,4)
To graph by hand find the values at 1 and 3 by plugging it into the original function and drawing a line between the two.
2(1)-2 = 0 Point (1,0)
2(3)-2 = 4 Point (3,4)
Compute the sum to simplify C :
Now, is the entry of the matrix in row i and column j, so
Answer:
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4
Step-by-step explanation:
The given function is presented as follows;
From the given function, we have;
When x = 1, the denominator of the fraction, , which is (x - 1) = 0, and the function becomes,
therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote
Also we have that in the given function, as <em>x</em> increases, the fraction tends to 0, therefore as x increases, we have;
Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1
When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, . When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left
Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.