Question

the graph of which function is decreasing over the interval $(-4, \infty)$?
\\(\circ\\) $f(x) = (x + 4)^2 + 4$
\\(\circ\\) $f(x) = -(x + 4)^2 + 4$
\\(\circ\\) $f(x) = (x - 4)^2 - 4$
\\(\circ\\) $f(x) = -(x - 4)^2 - 4$

Explanation:

Step1: Recall the vertex form of a parabola

The vertex form of a quadratic function is \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex and \(a\) determines the direction (upward if \(a>0\), downward if \(a<0\)) and the width. The axis of symmetry is \(x = h\).

Step2: Analyze each function

• For \( f(x)=(x + 4)^2+4 \): Here, \(a = 1>0\), vertex at \((-4, 4)\), axis of symmetry \(x=-4\). Since \(a>0\), the parabola opens upward. So, it is decreasing to the left of \(x = -4\) and increasing to the right of \(x=-4\) (i.e., on \((-4,\infty)\) it is increasing).
• For \( f(x)=-(x + 4)^2+4 \): Here, \(a=- 1<0\), vertex at \((-4, 4)\), axis of symmetry \(x = -4\). Since \(a<0\), the parabola opens downward. So, it is increasing to the left of \(x=-4\) and decreasing to the right of \(x = -4\) (i.e., on \((-4,\infty)\) it is decreasing).
• For \( f(x)=(x - 4)^2-4 \): Here, \(a = 1>0\), vertex at \((4,-4)\), axis of symmetry \(x = 4\). The interval \((-4,\infty)\) includes \(x = 4\), and the behavior around \(x = 4\) is different from what we need. Also, since \(a>0\), it opens upward, decreasing to the left of \(x = 4\) and increasing to the right of \(x = 4\), but the interval starts at \(-4\), so it is not decreasing on the entire \((-4,\infty)\).
• For \( f(x)=-(x - 4)^2-4 \): Here, \(a=-1<0\), vertex at \((4,-4)\), axis of symmetry \(x = 4\). Since \(a<0\), it opens downward, increasing to the left of \(x = 4\) and decreasing to the right of \(x = 4\), so it is not decreasing on \((-4,\infty)\).

Answer:

\( f(x)=-(x + 4)^2+4 \) (the second option)