Question

which is an equation for the parabola that has vertex (6,-3) and passes through the point (2,-2)?
a. \\( x = -4y^2 - 24y - 30 \\)
b. \\( x = -4y^2 - 24y - 42 \\)
c. \\( x = -16y^2 - 48y - 138 \\)
d. \\( x = 4y^2 - 24y + 42 \\)

Explanation:

Step1: Recall the vertex form of a horizontal parabola

The vertex form of a horizontal parabola (opening left or right) is \(x = a(y - k)^2 + h\), where \((h, k)\) is the vertex. Here, the vertex is \((6, -3)\), so \(h = 6\) and \(k = -3\). Substituting these into the formula, we get \(x = a(y + 3)^2 + 6\).

Step2: Substitute the point \((2, -2)\) to find \(a\)

We know the parabola passes through \((2, -2)\), so substitute \(x = 2\) and \(y = -2\) into the equation \(x = a(y + 3)^2 + 6\):
\[
2 = a(-2 + 3)^2 + 6
\]
Simplify the right - hand side: \(-2 + 3=1\), so the equation becomes \(2=a(1)^2 + 6\), which is \(2=a + 6\).

Step3: Solve for \(a\)

Subtract 6 from both sides of the equation \(2=a + 6\): \(a=2 - 6=-4\).

Step4: Write the equation of the parabola

Now that we know \(a=-4\), \(h = 6\), and \(k=-3\), the equation of the parabola in vertex form is \(x=-4(y + 3)^2+6\).

Step5: Expand the equation

First, expand \((y + 3)^2\) using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = y\) and \(b = 3\). So \((y + 3)^2=y^{2}+6y + 9\).
Then, multiply by \(-4\): \(-4(y^{2}+6y + 9)=-4y^{2}-24y-36\).
Add 6 to the result: \(x=-4y^{2}-24y-36 + 6=-4y^{2}-24y-30\).

Answer:

A. \(x = - 4y^{2}-24y - 30\)