Find the equation of a plane that is perpendicular to the vector −4i⃗ −4j⃗ −k⃗ and passing through the point (−2,−5,5)
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Problem 5
The function is continuous for the given domain
This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get
y = (-5/6)x+5
y = (-5/6)*6 + 5
y = -5+5
y = 0
This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.
The range is therefore
In interval notation, you can write the range as . The square bracket indicates "include this endpoint as part of the range".
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Problem 6
The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.
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If you plugged in x = -4, then you should get...
y = (-1/2)*(-4)+2
y = 2+2
y = 4
So the input x = -4 lead the output y = 4
Repeat for x = -2
y = (-1/2)x+2
y = (-1/2)*(-2)+2
y = 1+2
y = 3
and the same for x = 0 as well
y = (-1/2)x+2
y = (-1/2)*0 + 2
y = 0 + 2
y = 2
and x = 2 also
y = (-1/2)x+2
y = (-1/2)*2 + 2
y = -1+2
y = 1
Finally, plug in x = 4
y = (-1/2)x+2
y = (-1/2)*4+2
y = -2+2
y = 0
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If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}
Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}