Question

which function has two x-intercepts, one at (0, 0) and one at (4, 0)?
\\( f(x) = x(x - 4) \\)
\\( f(x) = x(x + 4) \\)
\\( f(x) = (x - 4)(x - 4) \\)
\\( f(x) = (x + 4)(x + 4) \\)

Explanation:

Step1: Recall x - intercept formula

To find the x - intercepts of a function \(y = f(x)\), we set \(y=0\) (i.e., \(f(x) = 0\)) and solve for \(x\). If a function is in factored form \(f(x)=(x - a)(x - b)\), then the x - intercepts are at \(x=a\) and \(x = b\) (when \(f(x)=0\)).

Step2: Analyze each option

• For \(f(x)=x(x - 4)\):

Set \(f(x)=0\), then \(x(x - 4)=0\). By the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So \(x=0\) or \(x - 4=0\Rightarrow x = 4\). The x - intercepts are \((0,0)\) and \((4,0)\).

• For \(f(x)=x(x + 4)\):

Set \(f(x)=0\), then \(x(x + 4)=0\). So \(x = 0\) or \(x+4=0\Rightarrow x=- 4\). The x - intercepts are \((0,0)\) and \((-4,0)\), not \((4,0)\).

• For \(f(x)=(x - 4)(x - 4)\):

Set \(f(x)=0\), then \((x - 4)(x - 4)=0\). So \(x-4 = 0\Rightarrow x = 4\) (a repeated root). The x - intercept is \((4,0)\) (with multiplicity 2), not \((0,0)\) and \((4,0)\) as two distinct intercepts in the way we need.

• For \(f(x)=(x + 4)(x + 4)\):

Set \(f(x)=0\), then \((x + 4)(x + 4)=0\). So \(x+4=0\Rightarrow x=-4\) (a repeated root). The x - intercept is \((-4,0)\), not what we want.

Answer:

A. \(f(x)=x(x - 4)\)