Asked by htht
How would you describe the relationship between the real zero(s) and x-intercept(s) of the function
When you set the function equal to zero, the solution is x = 1; therefore, the graph has an x-intercept of (1, 0).
When you set the function equal to zero, the solutions are x = 0 or x = 1; therefore, the graph has x-intercepts at (0, 0) and (1, 0).
When you substitute x = 0 into the function, there is no solution; therefore, the graph will not have any x-intercepts.
Since there are asymptotes at x = –3, x = –1, and x = 0, the graph has no x-intercepts and, therefore, no real zeros.
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GPT-5 mini
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Short answer: A real zero of a function is an x-value a for which f(a) = 0. Graphically that corresponds to an x‑intercept at (a,0), provided the function is defined at a.
Important caveats and details:
- If f(a) = 0 and f is defined at a, then the graph has the x‑intercept (a,0).
- If f(a) is undefined (hole or vertical asymptote) then a is not an x‑intercept even if algebraically the numerator would be zero there (common for rational functions). For rational functions, zeros occur where the numerator = 0 and the denominator ≠ 0.
- Multiplicity matters: a zero of odd multiplicity typically makes the graph cross the x‑axis; even multiplicity typically makes it touch and turn around at the axis.
- If the equation f(x) = 0 has no real solutions, the graph has no x‑intercepts (no real zeros).
Comments on the four example sentences:
1. Correct: if solving f(x)=0 gives x = 1 and f is defined at 1, the graph has intercept (1,0).
2. Correct: if f(x)=0 gives x = 0 or x = 1 (and f is defined at those x), the intercepts are (0,0) and (1,0).
3. Ambiguous as written: “substitute x = 0 into the function, there is no solution” suggests either f(0) ≠ 0 or f(0) is undefined. If f(0) ≠ 0 then indeed x=0 is not an x‑intercept; if f(0) is undefined you also do not have an x‑intercept at x=0. But absence of an intercept at x=0 does not rule out intercepts at other x.
4. Incorrect as a general statement: having vertical asymptotes at x = −3, −1, and 0 does not by itself guarantee there are no x‑intercepts anywhere — it only means the graph cannot have intercepts at those x values. There could still be x‑intercepts at other x where the function is defined.
Important caveats and details:
- If f(a) = 0 and f is defined at a, then the graph has the x‑intercept (a,0).
- If f(a) is undefined (hole or vertical asymptote) then a is not an x‑intercept even if algebraically the numerator would be zero there (common for rational functions). For rational functions, zeros occur where the numerator = 0 and the denominator ≠ 0.
- Multiplicity matters: a zero of odd multiplicity typically makes the graph cross the x‑axis; even multiplicity typically makes it touch and turn around at the axis.
- If the equation f(x) = 0 has no real solutions, the graph has no x‑intercepts (no real zeros).
Comments on the four example sentences:
1. Correct: if solving f(x)=0 gives x = 1 and f is defined at 1, the graph has intercept (1,0).
2. Correct: if f(x)=0 gives x = 0 or x = 1 (and f is defined at those x), the intercepts are (0,0) and (1,0).
3. Ambiguous as written: “substitute x = 0 into the function, there is no solution” suggests either f(0) ≠ 0 or f(0) is undefined. If f(0) ≠ 0 then indeed x=0 is not an x‑intercept; if f(0) is undefined you also do not have an x‑intercept at x=0. But absence of an intercept at x=0 does not rule out intercepts at other x.
4. Incorrect as a general statement: having vertical asymptotes at x = −3, −1, and 0 does not by itself guarantee there are no x‑intercepts anywhere — it only means the graph cannot have intercepts at those x values. There could still be x‑intercepts at other x where the function is defined.
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