Question

Help check and answer
medal + fan!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

Part A: Explain why the x-coordinates of the points where the graphs of the equations y = 4^–x and y = 2x +^ 3 intersect are the solutions of the equation 4–x = 2x + 3. (4 points)

Part B: Make tables to find the solution to 46–x = 2x + ^3. Take the integer values of x between −3 and 3. (4 points)

Part C: How can you solve the equation 4^–x = 2x + ^3 graphically? (2 points)

Answer 2

@HolsterEmission

Answer 3

Here is what i have so far

Answer 4

Part A:
Where the two lines intersect is the answer to both solutions because it is a point for both equations.
Part B:??????????????
Part C: You would draw the lines on the graph and the solution will be the point where the two lines intersect.

Answer 5

is it correct???
And what is part B?

Answer 6

@HolsterEmission

Answer 7

I'm confused... what are the two equations? They look like \(y=4^{-x}\) and \(y=2x+3\) but they keep getting minor changes wherever they're replicated.

Answer 8

What is "+^"?

Answer 9

Right, clearly something is being raised to some power, but *what* is being raised?

Answer 10

Still not answering my question... Is the second equation \(2x^3\)? Because that's not what you've been saying.

Answer 11

Yes, immensely. For future reference, if you'd like to communicate something of the second equation's form, try writing it as "base^(exponent)" with parentheses around the entire exponent, like 2^(x+3). "+ ^3" is not clear at all.

As for the actual question: I don't think your answer for the first part is adequate. You should be more specific about what it means for an intersection to occur. The intersection is some point \((x,y)\) where both functions take on the same values of \(x\) and \(y\). In other words, you're setting up a point equation that says \((4^{-x},y)=(2^{x+3},y)\), which reduces to the single equation \(4^{-x}=2^{x+3}\) because you want to guarantee that both points have the same value as the \(y\) coordinate.

The second part asks you to fill in the following table:
\[\begin{array}{c|ccccccc}
x&-3&-2&-1&0&1&2&3\\[1ex]
\hline
y=4^{-x}&\\[1ex]
y=2^{x+3}&
\end{array}\]where filling the bottom two rows is just a matter of finding the value of \(y\) for each given value of \(x\).

Your answer for part C is fine.

Answer 12

is it 0.16?

Answer 13

\[\begin{array}{c|ccccccc}
x&-3&-2&-1&0&1&2&3\\[1ex]
\hline
y=4^{-x}&\\[1ex]
y=2^{x+3}&
\end{array}\]When \(x=-3\), the first equation gives
\[y=4^{-(-3)}=4^3=64\]In case you were asking about the case when \(x=3\), you have
\[y=4^{-3}=\frac{1}{4^3}=\frac{1}{64}=0.015625\]which you could round to \(0.016\), but not \(0.16\).

Answer 14

ok thanks!

Answer 15

what would I write for part A?

Answer 16

? @HolsterEmission

Answer 17

This good?
Part A:
Where the two lines intersect is the answer to both solutions because it is a point for both y = 4^–x and y = 2x + 3.

Part B:
x y = 4^–x y = 2x + 3
-3 y= 64 y=-3
-2 y=16 y=-1
-1 y=4 y=1
0 y= 1 y=3
1 y=0.25 y=5
2 y=0.0625 y=7
3 y=0.015625 y=9

Part C: You would draw the lines on the graph and the solution will be the point where the two lines intersect.