The third graph represents a function.
In a function, every input (x value) has <em>exactly</em> one output (y value). If even a single input has zero or two outputs, the graph does not represent a function.
A good way of testing this is using a vertical line. As you move a vertical line from left to right across a graph, it should always be touching exactly one point on the graphed line.
In this case, every graph fails this vertical line test except for the third graph, so the third graph represents a function.
For this case what you should do is evaluate values of x in the function and verify that they meet the result of f (x) shown in the graph.
The answer is
f (x) = - 2lxl +1
notice that
f (1) = - 2l1l + 1 = -1
f (-1) = - 2l-1l + 1 = -1
Both comply with the value of f (x) shown in the graph
answer
f (x) = - 2lxl +1
<span>Find the equation of the line parallel to the line y = 4x – 2 that passes through the point (–1, 5).
</span>y = 4x – 2 has slope = 4
<span>parallel lines have same slope so slope = 4
</span><span>passes through the point (–1, 5).
</span><span>y = mx+b
5 = 4(-1) + b
b =9
equation
y = 4x + 9
answer
The slope of y = 4x – 2 is 4
The slope of a line parallel to y = 4x – 2 is 4
The equation of the line parallel to y = 4x – 2 that passes through the point (–1, 5) is y = 4x + 9</span>
9514 1404 393
Answer:
B) f(x) = 3
Step-by-step explanation:
The value in the f(x) column is always 3, so the only reasonable choice is ...
f(x) = 3
Answer:
Acute
Step-by-step explanation:
This would not form a triangle because the lengths are too long. If it did, the triangle would be acute.
