Which function has an asymptote at x = pi?

Mathematics

1 answer:

Ber [7]2 years ago

3 0

Answer:

C. y = csc x

Step-by-step explanation:

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Idently the graph of f(x) = 3(x - 1) - 4. Then identity the vertex and ads of symmetry find the minimum value of and describe wh

Vesna [10]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x) = 3(x-1)^2-4$

The equation is already in vertex form $f(x)=a(x-h)^2+k$ where $a$ affects how "fat" or "skinny" the parabola is and $(h,k)$ is the vertex. Therefore, the vertex is $(h,k)\rightarrow(1,-4)$ .

The axis of symmetry is a line where the parabola is cut into two congruent halves. This is defined as $x=h$ for a parabola with a vertical axis. Hence, the axis of symmetry is $x=1$ .

The minimum value is the smallest value in the range of the function. In the case of a parabola, the y-coordinate of the vertex is the minimum value. Therefore, the minimum value is $y=-4$ .