The graph shown here is the graph of which of the following rational functions?
1 answer:
Considering it's asymptotes, the rational function shown in the graph is:
C.
- The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.
- The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity.
In this graph, there are vertical asymptotes at x = -2 and x = 1, hence the denominator of the function is:
(x + 2)(x - 1).
The horizontal asymptote is of y = 0, hence the function can be given by:
C.
More can be learned about rational functions and asymptotes at brainly.com/question/24410059
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Ok, so when it says "3x, x2," it means that when the x value is less than or equal to 3, you graph 3x. You would have the following points on the line: (-3,-9), (-2,-6), (-1,-3), (0,0), (1,3), and (2,6). The endpoint of the ray on the left side (the points I just said) would be (2,6)
When it says, "1/2 x +5, 2x," it means that when the value of x is greater than or equal to 2, the y value is .5x + 5. You would graph the following points: (2,6), (4,7), (6,8), and (8,9). The endpoint of th ray on the right (the points I just said) would be (2,6).
Keep in mind that though the graph is continuous, it isn't one line, but 2 rays (which you probably already know).
We can use the vertex form
to find the equation of the graph. We know that a is the coefficient, and since the graph opens upwards and is slightly stretched vertically, we know that a is 1/3. On the graph, we see that the vertex of the parabola is at (0,-1). Therefore, h is 0 and k is -1.
Our vertex form equation is
Which is D
to find the equation of the graph. We know that a is the coefficient, and since the graph opens upwards and is slightly stretched vertically, we know that a is 1/3. On the graph, we see that the vertex of the parabola is at (0,-1). Therefore, h is 0 and k is -1.
Our vertex form equation is
Which is D