Asked by htht
Use the graph of f(x) to explain why the function has no real zeros.
The graph of f(x) does not pass through (0, 0); therefore, f(x) has no real zeros.
The graph of f(x) does not cross the x-axis; therefore, f(x) has no real zeros.
The graph of f(x) does not cross the y-axis; therefore, f(x) has no real zeros.
The graph of f(x) has one or more asymptotes; therefore, f(x) has no real zeros.
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Best choice: "The graph of f(x) does not cross the x-axis; therefore, f(x) has no real zeros."
Explanation: a real zero is an x-value where f(x)=0, i.e. a point where the graph intersects the x-axis. If the graph has no x-intercepts, then there are no real zeros. (More precisely, say "does not intersect the x-axis" — if a graph merely touches the x-axis it still gives a real zero.)
Why the others are wrong:
- Not passing through (0,0) only rules out x=0 as a zero; it says nothing about other x-values.
- Not crossing the y-axis is irrelevant (y-axis intersections are values at x=0).
- Having asymptotes does not by itself prohibit intersections with the x-axis.
Explanation: a real zero is an x-value where f(x)=0, i.e. a point where the graph intersects the x-axis. If the graph has no x-intercepts, then there are no real zeros. (More precisely, say "does not intersect the x-axis" — if a graph merely touches the x-axis it still gives a real zero.)
Why the others are wrong:
- Not passing through (0,0) only rules out x=0 as a zero; it says nothing about other x-values.
- Not crossing the y-axis is irrelevant (y-axis intersections are values at x=0).
- Having asymptotes does not by itself prohibit intersections with the x-axis.
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