Choose the point-slope form of the equation below that represents the line that passes through the points (-6, 4) and (2, 0).

1 answer:

5 0

Point slope form
y-y1=m(x-x1)
slope=m
a point is (x1,y1)

slope=(y2-y1)/(x2-x1)
given
(-6,4) and (2,0)
slope=(0-4)/(2-(-6))=-4/(2+6)=-4/8=-1/2

a point is (-6,4)
x1=-6
y1=4

y-4=-1/2(x+6)
not listed

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Which is the interval notation represent the domain?

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Answer:

The first choice, $(-\infty, 7) \cup (7, \infty)$ .

Step-by-step explanation:

The dashed vertical line is a vertical asymptote. The x-coordinate of all points on this vertical asymptote are equal to 7. In other words, when the x-value of a point on the graph approaches $7$ , its y-value approaches infinity (or negative infinity.)

Either way, the graph is not defined for $x = 7$ . The point $7$ should thus be excluded from the domain of the graph.

The graph is apparently defined for all other x-values. The domain of the function should thus be all real numbers with the exception of $x = 7$ . Here's how to write that in interval notation:

The set of all real numbers can be expressed as the interval $(-\infty, \infty)$ . Note that infinity $\infty$ (or $(- \infty)$ itself isn't a real number. The round brackets indicate that both endpoints are excluded from the from the interval.
The set of all real numbers less than (not equal to) $7$ is $(-\infty,7 )$ (both ends are excluded.) The set of all real numbers greater than (not equal to) $7$ is $(7, \infty)$ .
The set of all real numbers that is not equal to $7$ is the union of all real numbers less than $7$ and all real numbers greater than $7$ . The union operator $\cup$ connects two intervals.