Question

question 24 (1 point)
(02.08 mc)
what is the equation of the quadratic graph with a focus of (3, -1) and a directrix of y = 1?
o a
f(x)=-\frac{1}{4}(x - 3)^2+1
o b
f(x)=-\frac{1}{4}(x - 3)^2
o c
f(x)=\frac{1}{4}(x - 3)^2+1
o d
f(x)=\frac{1}{4}(x - 3)^2

Explanation:

Step1: Recall the formula for parabola

The standard - form of a parabola with a vertical axis of symmetry is \((x - h)^2=4p(y - k)\), where \((h,k)\) is the vertex and \(p\) is the distance between the vertex and the focus (or the vertex and the directrix). The vertex is the mid - point between the focus \((x_f,y_f)=(3, - 1)\) and a point on the directrix. Since the directrix is \(y = 1\), the \(x\) - coordinate of the vertex \(h\) is the same as the \(x\) - coordinate of the focus, so \(h = 3\). The \(y\) - coordinate of the vertex \(k=\frac{y_f + y_d}{2}\), where \(y_f=-1\) and \(y_d = 1\). Then \(k=\frac{-1 + 1}{2}=0\).

Step2: Calculate the value of \(p\)

The distance \(p\) between the vertex \((3,0)\) and the focus \((3,-1)\) is \(p=-1\) (negative because the focus is below the vertex).

Step3: Write the equation of the parabola

Substitute \(h = 3\), \(k = 0\), and \(p=-1\) into the formula \((x - h)^2=4p(y - k)\). We get \((x - 3)^2=4\times(-1)(y-0)\), which simplifies to \(y=-\frac{1}{4}(x - 3)^2\).

Answer:

B. \(f(x)=-\frac{1}{4}(x - 3)^2\)