Question

find the vertex, axis of symmetry, x-intercept, y-intercepts, focus, and directrix for the parabola. sketch the graph.
x = \frac{1}{2}(y - 4)^2 - 5

the vertex is \square.
(simplify your answer. type an ordered pair.)

the axis of symmetry is \square = \square.
(simplify your answer.)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\bigcirc a. the x-intercept(s) is/are \square.
\\ \\ \\ \\ (simplify your answer. type an ordered pair. type an exact answer, using radicals as
\\ \\ \\ \\ needed. use a comma to separate answers as needed.)
\bigcirc b. there are no x-intercepts.

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\bigcirc a. the y-intercept(s) is/are \square.
\\ \\ \\ \\ (simplify your answer. type an ordered pair. type an exact answer, using radicals as
\\ \\ \\ \\ needed. use a comma to separate answers as needed.)
\bigcirc b. there are no y-intercepts.

the focus is \square.
(simplify your answer. type an ordered pair.)

the directrix is \square = \square.
(simplify your answer.)

use the graphing tool to graph the parabola.

Explanation:

Step1: Find the Vertex

The standard form of a parabola that opens horizontally is \( x = a(y - k)^2 + h \), where \((h, k)\) is the vertex. For the given equation \( x=\frac{1}{2}(y - 4)^2-5 \), we can identify \( h=-5 \) and \( k = 4 \). So the vertex is \((-5, 4)\).

Step2: Find the Axis of Symmetry

For a horizontal parabola \( x = a(y - k)^2 + h \), the axis of symmetry is the horizontal line \( y=k \). Here, \( k = 4 \), so the axis of symmetry is \( y = 4 \).

Step3: Find the x - intercept

To find the x - intercept, set \( y = 0 \) in the equation \( x=\frac{1}{2}(y - 4)^2-5 \).
Substitute \( y = 0 \):
\( x=\frac{1}{2}(0 - 4)^2-5=\frac{1}{2}(16)-5=8 - 5 = 3 \)
So the x - intercept is \((3,0)\), so we choose option A and the x - intercept is \((3,0)\).

Step4: Find the y - intercept

To find the y - intercept, set \( x = 0 \) in the equation \( x=\frac{1}{2}(y - 4)^2-5 \).
\( 0=\frac{1}{2}(y - 4)^2-5 \)
Add 5 to both sides: \( 5=\frac{1}{2}(y - 4)^2 \)
Multiply both sides by 2: \( 10=(y - 4)^2 \)
Take square roots: \( y-4=\pm\sqrt{10} \)
\( y = 4\pm\sqrt{10} \)
So the y - intercepts are \((0,4+\sqrt{10})\) and \((0,4 - \sqrt{10})\), so we choose option A.

Step5: Find the Focus

For a horizontal parabola \( x = a(y - k)^2 + h \), the focus is at \((h+\frac{1}{4a},k)\). Here, \( a=\frac{1}{2} \), \( h=-5 \), \( k = 4 \).
First, calculate \( \frac{1}{4a}=\frac{1}{4\times\frac{1}{2}}=\frac{1}{2} \)
Then \( h+\frac{1}{4a}=-5+\frac{1}{2}=-\frac{10 - 1}{2}=-\frac{9}{2}=-4.5 \)
So the focus is \((-\frac{9}{2},4)\) or \((-4.5,4)\).

Step6: Find the Directrix

For a horizontal parabola \( x = a(y - k)^2 + h \), the directrix is the vertical line \( x=h-\frac{1}{4a} \).
We know \( h=-5 \) and \( \frac{1}{4a}=\frac{1}{2} \)
So \( x=-5-\frac{1}{2}=-\frac{10 + 1}{2}=-\frac{11}{2}=-5.5 \)
So the directrix is \( x=-\frac{11}{2} \)

Answer:

The vertex is \(\boldsymbol{(-5, 4)}\).
The axis of symmetry is \(\boldsymbol{y = 4}\).
For x - intercept: A. The x - intercept(s) is/are \(\boldsymbol{(3,0)}\).
For y - intercept: A. The y - intercept(s) is/are \(\boldsymbol{(0,4+\sqrt{10}),(0,4 - \sqrt{10})}\).
The focus is \(\boldsymbol{(-\frac{9}{2},4)}\) (or \((-4.5,4)\)).
The directrix is \(\boldsymbol{x=-\frac{11}{2}}\) (or \(x=-5.5\)).