if f(x) is a linear function and the domain of f(x) is the set of all real numbers, which statement cannot…

if f(x) is a linear function and the domain of f(x) is the set of all real numbers, which statement cannot be true?\nthe graph of f(x) has zero x - intercepts.\nthe graph of f(x) has exactly one x - intercept.\nthe graph of f(x) has exactly two x - intercepts.\nthe graph of f(x) has infinitely many x - intercepts.

if f(x) is a linear function and the domain of f(x) is the set of all real numbers, which statement cannot be true?\nthe graph of f(x) has zero x - intercepts.\nthe graph of f(x) has exactly one x - intercept.\nthe graph of f(x) has exactly two x - intercepts.\nthe graph of f(x) has infinitely many x - intercepts.

Answer

Explanation:

Step1: 線形関数の一般形を考える

線形関数は $y = mx + b$（$m,b$ は定数、$m\neq0$ のとき直線、$m = 0$ のとき定数関数）の形で表される。$x$ 切片は $y = 0$ のときの $x$ の値で、$0=mx + b$ を解くと $x=-\frac{b}{m}(m\neq0)$ である。$m = 0$ のとき、$y=b$。

Step2: $x$ 切片の個数を分析する

  • $m\neq0$ のとき、$0 = mx + b$ は1つの解を持つので、$x$ 切片は1つ。
  • $m = 0,b\neq0$ のとき、$y = b\neq0$ で、$x$ 軸と交わらないので $x$ 切片は0つ。
  • $m = 0,b = 0$ のとき、$y = 0$ で、$x$ 軸全体がグラフと一致し、無数の $x$ 切片を持つ。
  • 線形関数のグラフは直線または定数関数のグラフであり、2つの $x$ 切片を持つことはできない。

Answer:

The graph of $f(x)$ has exactly two $x$-intercepts.