Two perpendicular lines intersect on the y - axis. If the equation of the first line is x + y = 1, what is the equation of the s
1 answer:
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Remark
The very first thing you should do with a question like this is get the graph. Then you will know what you are looking for. I have provided you with such a graph. Your table should center around -2 ≤ x ≤ -1
So let's set up a table and see what we get. Start with y = x + 5
x x + 5
-1 4
-1.25 3.75
-1.5 3.5
-1.75 3.25
-2 3
Do the same thing for - (3)^x + 4 See below to see how this is entered your calculator
x -(3)^x + 4
-1 3.66
-1.25 3.75
-1.5 3.8
-1.75 3.85
-2 3.89
Conclusion
When x = - 1.25 y = 3.75 for both graphs. <<<< Answer
Footnote
You may not be familiar with how to put this in your calculator. This is the way I would do it. I'm only doing it for y = -(3^x) + 4
Let x = - 1.25
3
^ Note your calculator might have x^y or y^x. You'll have 1 of the three.
1.25
+/-
=
X
1
+/-
= At this point you should have -0.25
+
4
=
That gives you 3.75
The very first thing you should do with a question like this is get the graph. Then you will know what you are looking for. I have provided you with such a graph. Your table should center around -2 ≤ x ≤ -1
So let's set up a table and see what we get. Start with y = x + 5
x x + 5
-1 4
-1.25 3.75
-1.5 3.5
-1.75 3.25
-2 3
Do the same thing for - (3)^x + 4 See below to see how this is entered your calculator
x -(3)^x + 4
-1 3.66
-1.25 3.75
-1.5 3.8
-1.75 3.85
-2 3.89
Conclusion
When x = - 1.25 y = 3.75 for both graphs. <<<< Answer
Footnote
You may not be familiar with how to put this in your calculator. This is the way I would do it. I'm only doing it for y = -(3^x) + 4
Let x = - 1.25
3
^ Note your calculator might have x^y or y^x. You'll have 1 of the three.
1.25
+/-
=
X
1
+/-
= At this point you should have -0.25
+
4
=
That gives you 3.75
From the odd-degree terms, take out one copy and rewrite the series as
Then if |<em>x</em>| < 1, we can condense this to
Since the series we invoked here converge on -1 < <em>x</em> < 1, so does this one.