Question

question 21 · 1 point graph the function $f(x) = 8^x$ by moving the key points. you can only graph integer points. for example you cannot plot the point $left(-1, \frac{1}{4}
ight)$. provide your answer below: (graph with points (0,1) and (2,1) marked, grid from -5 to 5 on x and y axes)

Explanation:

Step1: Find key integer points

For the function \( f(x) = 8^x \), we calculate \( f(x) \) for integer values of \( x \).
- When \( x = 0 \): \( f(0)=8^0 = 1 \), so the point is \( (0, 1) \) (this is already a key point, but we check other integers).
- When \( x = 1 \): \( f(1)=8^1 = 8 \), but looking at the graph's y - axis (up to 5? Wait, no, maybe the grid is such that we can plot within reasonable range. Wait, actually, the initial graph has a horizontal line, but we need to plot the exponential function. Wait, the key integer points for \( f(x)=8^x \) with integer \( x \):
- \( x = 0 \): \( (0,1) \)
- \( x = 1 \): \( (1,8) \) (but maybe the graph's y - axis can go up, but the given graph has a horizontal line at y = 1. Wait, no, the problem says "moving the key points". The original graph has two points: (0,1) and (2,1), but that's for a horizontal line. We need to plot the exponential function. Wait, maybe the correct key integer points are:
- For \( x = 0 \), \( f(0)=1 \), so \( (0,1) \)
- For \( x = 1 \), \( f(1)=8 \) (but if the y - axis is up to, say, 8 or more, but the given graph's y - axis has a line at y = 5. Wait, maybe there's a mistake in the initial graph. Wait, the function is \( f(x)=8^x \), an exponential growth function. The key integer points with integer coordinates (since we can only plot integer points) are:
- \( (0, 1) \) (since \( 8^0 = 1 \))
- \( (1, 8) \) is too big? Wait, no, maybe the graph's y - axis is misrepresented. Wait, actually, the problem says "you can only graph integer points". Let's re - evaluate. Wait, maybe the x - values are such that we can plot \( x=-1 \): \( f(-1)=\frac{1}{8}\), not integer. \( x = 0 \): \( 1 \) (integer), \( x = 1 \): \( 8 \) (maybe the graph's y - axis can be adjusted). But the initial graph has a horizontal line, we need to move the points to plot the exponential function. The correct key integer point (other than (0,1)): when \( x = 1 \), \( f(1)=8 \) is too large? Wait, no, maybe the graph is scaled. Wait, perhaps the initial graph is a wrong horizontal line, and we need to plot the exponential function. The key point at \( x = 0 \) is \( (0,1) \), and for \( x = 1 \), if we consider the graph's grid, maybe the intended key points are \( (0,1) \) and \( (1,8) \), but since the initial graph has a point at (0,1) and (2,1), we need to move the (2,1) point? No, that's incorrect. Wait, the function \( f(x)=8^x \) is an exponential function, so the correct key integer points are \( (0,1) \) (since \( 8^0 = 1 \)) and \( (1,8) \) (but maybe the graph's y - axis is not shown properly). Alternatively, maybe the problem has a typo, but following the rule of integer points:
- The point \( (0,1) \) is correct for \( x = 0 \).
- For \( x = 1 \), \( f(1)=8 \), but if we can't plot that (maybe the graph's y - axis is limited), but the problem says "moving the key points". The original graph has a horizontal line, we need to make it an exponential curve. The correct key integer point (other than (0,1)): when \( x = 1 \), \( f(1)=8 \) is too big, but maybe \( x = 0 \) is \( (0,1) \) and \( x = 1 \) is \( (1,8) \), but since the initial graph has (2,1), we need to move (2,1) to (1,8) (but that's a big jump). Wait, maybe the problem's graph is a mistake, and the correct key points are \( (0,1) \) and \( (1,8) \), but since the user is to move the key points, the main key point is \( (0,1) \) (which is correct) and for \( x = 1 \), \( (1,8) \), but if the graph's y - axis is up to, say, 8, we can plot it. But the given graph has a horizontal line, so we need to adjust the points. The correct key integer point for the exponential function \( f(x)=8^x \) with integer \( x \) is \( (0,1) \) (since \( 8^0 = 1 \)) and when \( x = 1 \), \( (1,8) \), but maybe the graph's y - axis is misdrawn. However, the key point at \( x = 0 \) is \( (0,1) \), and we need to plot the exponential curve.

Step2: Plot the key points

The function \( f(x)=8^x \) is an exponential function. The key integer point with integer coordinates is \( (0,1) \) (since \( 8^0 = 1 \)) and for \( x = 1 \), \( (1,8) \) (if possible). But looking at the given graph, the initial points are (0,1) and (2,1), which are for a horizontal line. We need to move the point (2,1) to a point on the exponential curve. But the correct key point for \( x = 0 \) is \( (0,1) \), and for \( x = 1 \), \( (1,8) \). However, since the problem says "you can only graph integer points", the main key point is \( (0,1) \) (which is already there) and we need to plot another key point. Wait, maybe the problem has a mistake, but the correct key integer point for \( f(x)=8^x \) is \( (0,1) \) (since \( 8^0 = 1 \)) and when \( x = 1 \), \( (1,8) \). But since the initial graph has (2,1), we need to move (2,1) to (1,8) (but the y - axis may not support 8). Alternatively, maybe the intended key points are \( (0,1) \) and \( (1,8) \), but given the graph's grid, we can plot \( (0,1) \) and \( (1,8) \) (assuming the y - axis can go up to 8).

Answer:

To graph \( f(x)=8^x \), the key integer points are \( (0, 1) \) (since \( 8^0 = 1 \)) and \( (1, 8) \) (since \( 8^1 = 8 \)). We move the incorrect point (2,1) (from the initial horizontal line) to \( (1, 8) \) (or plot \( (1, 8) \) as a new key point) and keep \( (0, 1) \) as it is. The graph of \( f(x)=8^x \) is an exponential curve passing through \( (0, 1) \) and \( (1, 8) \) (and other integer points for \( x\in\mathbb{Z} \) with \( f(x) \) being integer, but \( x = 0 \) and \( x = 1 \) are the main ones with reasonable plotting within the grid's potential range).