Question

select the correct answer.
a parabola has a focus and a directrix as shown on the graph.
what is the equation representing this parabola?
a. $f(x) = -\frac{1}{20}(x - 4)^2 - \frac{1}{2}$
b. $f(x) = -\frac{1}{10}(x - 4)^2 - 3$
c. $f(x) = -\frac{1}{20}(x - 4)^2 - 3$
d. $f(x) = -\frac{1}{10}(x - 4)^2 - \frac{1}{2}$

Explanation:

Step1: Identify vertex of parabola

The vertex \((h,k)\) of a parabola is halfway between the focus and the directrix. The focus is at \((4, -3)\) (from the graph) and the directrix is \(y = 2\). The midpoint (vertex) in \(y\)-coordinate: \(\frac{-3 + 2}{2}=\frac{-1}{2}\)? Wait, no, wait. Wait, the focus is at \((4, -3)\)? Wait, no, looking at the graph, the focus is at \((4, -3)\)? Wait, the directrix is \(y = 2\), and the focus is at \((4, -3)\). Wait, the vertex is halfway between focus and directrix. So the \(x\)-coordinate of vertex is same as focus, \(x = 4\). The \(y\)-coordinate: \(\frac{-3 + 2}{2}=\frac{-1}{2}\)? Wait, no, that can't be. Wait, maybe I misread the focus. Wait, the graph: the focus is at \((4, -3)\)? Wait, the grid: each square is 1 unit? Let's check again. The directrix is the red dashed line at \(y = 2\). The focus is a red dot at \((4, -3)\)? Wait, no, maybe the focus is at \((4, -3)\)? Wait, the vertex is halfway between focus \((4, k_f)\) and directrix \(y = d\). So \(k=\frac{k_f + d}{2}\). If directrix is \(y = 2\) and focus is at \((4, -3)\), then \(k=\frac{-3 + 2}{2}=-\frac{1}{2}\). Wait, but the vertex form of a parabola is \(f(x)=a(x - h)^2 + k\), where \((h,k)\) is vertex, and \(a=\frac{1}{4p}\), where \(p\) is the distance from vertex to focus (or directrix). The sign of \(a\) is negative because the parabola opens downward (focus is below directrix).

Wait, let's re-express. The standard form of a parabola that opens up or down is \((x - h)^2 = 4p(y - k)\), where \((h,k)\) is vertex, \(p\) is distance from vertex to focus (positive if opens up, negative if opens down). Here, directrix \(y = 2\), focus \((4, -3)\). So vertex \((h,k)\): \(h = 4\) (same \(x\) as focus), \(k\) is midpoint of focus \(y\) and directrix \(y\): \(k=\frac{-3 + 2}{2}=-\frac{1}{2}\). Then \(p\) is the distance from vertex to focus: \(p=-3 - (-\frac{1}{2})=-\frac{5}{2}\) (since focus is below vertex, \(p\) is negative). Then \(4p = 4*(-\frac{5}{2})=-10\). So the equation is \((x - 4)^2 = -10(y + \frac{1}{2})\). Solving for \(y\): \(y + \frac{1}{2}=-\frac{1}{10}(x - 4)^2\), so \(y=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\). Wait, but let's check the options. Option D: \(f(x)=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\). Wait, but maybe I made a mistake in the focus. Wait, maybe the focus is at \((4, -3)\)? Wait, no, let's check the grid again. The directrix is \(y = 2\), the focus is at \((4, -3)\)? Wait, the vertical distance between directrix (\(y=2\)) and focus (\(y=-3\)) is \(2 - (-3)=5\) units. So the vertex is halfway, so \(5/2 = 2.5\) units from each. So vertex \(y\)-coordinate: \(2 - 2.5=-0.5=-\frac{1}{2}\), which matches. Then \(p\) is the distance from vertex to focus: \(p=-3 - (-\frac{1}{2})=-\frac{5}{2}\). Then \(4p = -10\), so \(a=\frac{1}{4p}=-\frac{1}{10}\). So the equation is \(f(x)=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\), which is option D? Wait, no, wait the options:

Option A: \(-\frac{1}{20}(x - 4)^2-\frac{1}{2}\)

Option B: \(-\frac{1}{10}(x - 4)^2-3\)

Option C: \(-\frac{1}{20}(x - 4)^2-3\)

Option D: \(-\frac{1}{10}(x - 4)^2-\frac{1}{2}\)

Wait, so according to the calculation, \(a = -\frac{1}{10}\), \(h = 4\), \(k = -\frac{1}{2}\), so the equation is \(f(x)=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\), which is option D? Wait, but let's check again. Wait, maybe the focus is at \((4, -3)\), directrix \(y = 2\). The distance between focus and directrix is \(|2 - (-3)| = 5\), so \(p = -\frac{5}{2}\) (since it opens downward). Then \(4p = -10\), so \(a = \frac{1}{4p}=-\frac{1}{10}\). Vertex is at \((4, \frac{-3 + 2}{2})=(4, -\frac{1}{2})\). So the equation is \(y = -\frac{1}{10}(x - 4)^2 - \frac{1}{2}\), which is option D. Wait, but let's check the options again. Option D is \(f(x)=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\), which matches.

Wait, but maybe I made a mistake in the focus. Let's confirm the focus coordinates. From the graph, the red dot is at (4, -3)? Let's count the grid: x=4, y=-3 (since below the x-axis, 3 units down). Directrix is at y=2 (above x-axis, 2 units up). So yes, focus (4, -3), directrix y=2. Then vertex (4, ( -3 + 2)/2 )=(4, -0.5). Then \(p\) is the distance from vertex to focus: -3 - (-0.5)= -2.5= -5/2. Then \(4p= -10\), so \(a=1/(4p)= -1/10\). So the equation is \(y= -1/10 (x - 4)^2 - 1/2\), which is option D.

Step2: Check the value of \(a\) and vertex

The vertex is \((4, -\frac{1}{2})\), so the equation should have \((x - 4)^2\) and \(+ (-\frac{1}{2})\) at the end. The coefficient \(a\) is \(-\frac{1}{10}\) because \(4p = -10\) (from \(p = -\frac{5}{2}\)). So the equation is \(f(x)=-\frac{1}{10}(x - 4)^2-\frac{1}{2}\), which is option D.

Answer:

D. \( f(x) = -\frac{1}{10}(x - 4)^2 - \frac{1}{2} \)