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To graph a horizontal parabola of the form $$x = ay^2 + by + c$$ or $$x = a(y - k)^2 + h$$ using its properties, follow a six-step procedure:

Step 1. Determine whether the parabola opens to the left or to the right.
Step 2. Find the axis of symmetry.
Step 3. Find the vertex.
Step 4. Find the $$x$$-intercept. Find the point symmetric to the $$x$$-intercept across the axis of symmetry.
Step 5. Find the $$y$$-intercepts.
Step 6. Graph the parabola.

Graphing a Horizontal Parabola Using Properties

To graph the horizontal parabola $$x = 2y^2$$ using its properties, follow these steps:
Since the coefficient $$a = 2$$ is positive, the parabola opens to the right.
To find the axis of symmetry, use $$y = -\frac{b}{2a}$$. Since there is no $$y$$ term, $$b = 0$$, so the axis of symmetry is the horizontal line $$y = 0$$ (the $$x$$-axis).
The vertex lies on the axis of symmetry. Substituting $$y = 0$$ into the equation gives $$x = 2(0)^2 = 0$$, so the vertex is $$(0, 0)$$.
Because the vertex is at the origin, both the $$x$$- and $$y$$-intercepts are at the point $$(0, 0)$$.
To draw the parabola, select additional values for $$y$$ to find more points. For instance, when $$y = 1$$, $$x = 2(1)^2 = 2$$, giving the point $$(2, 1)$$. When $$y = 2$$, $$x = 2(2)^2 = 8$$, giving the point $$(8, 2)$$.
Next, plot the points symmetric to these across the axis of symmetry ($$y = 0$$). These symmetric points are $$(2, -1)$$ and $$(8, -2)$$.
Finally, plot the vertex and these points, and connect them with a smooth curve opening to the right.

Graphing $$x = 2y^2$$

To graph the horizontal parabola $$x = -4(y+1)^2 + 4$$ using its properties, follow these steps: First, identify the constants $$a$$, $$h$$, and $$k$$. Here, $$a = -4$$, $$h = 4$$, and $$k = -1$$. Since $$a = -4$$, the parabola opens to the left. The axis of symmetry is $$y = k$$, which is $$y = -1$$. The vertex is $$(h, k)$$, which is $$(4, -1)$$. Find the $$x$$-intercept by substituting $$y = 0$$ into the equation: $$x = -4(0+1)^2 + 4 = 0$$, so the $$x$$-intercept is $$(0, 0)$$. The point symmetric to $$(0, 0)$$ across the axis of symmetry is $$(0, -2)$$. Find the $$y$$-intercepts by substituting $$x = 0$$ into the equation: $$0 = -4(y+1)^2 + 4$$, which simplifies to $$-4 = -4(y+1)^2$$, and then $$1 = (y+1)^2$$. Solving $$y+1 = \pm 1$$ gives $$y = 0$$ and $$y = -2$$. The $$y$$-intercepts are $$(0, 0)$$ and $$(0, -2)$$. Finally, graph the parabola using the vertex, axis of symmetry, and intercept points.

Graphing $$x = -4(y+1)^2 + 4$$

To graph the horizontal parabola $$x = -4(y+2)^2 + 4$$ using its properties, first identify the constants $$a$$, $$h$$, and $$k$$. Here, $$a = -4$$, $$h = 4$$, and $$k = -2$$. Since $$a$$ is negative, the parabola opens to the left. The axis of symmetry is the horizontal line $$y = -2$$. The vertex is $$(4, -2)$$. To find the $$x$$-intercept, substitute $$y = 0$$: $$x = -4(0+2)^2 + 4 = -4(4) + 4 = -12$$. The $$x$$-intercept is $$(-12, 0)$$. The point symmetric to $$(-12, 0)$$ across the axis of symmetry is $$(-12, -4)$$. To find the $$y$$-intercepts, set $$x = 0$$: $$0 = -4(y+2)^2 + 4$$, which gives $$1 = (y+2)^2$$. Solving yields $$y+2 = \pm 1$$, so $$y = -1$$ and $$y = -3$$. The $$y$$-intercepts are $$(0, -1)$$ and $$(0, -3)$$. Plot these points to complete the graph.

Graphing $$x = -4(y+2)^2 + 4$$

To graph the horizontal parabola $$x = -2(y+3)^2 + 2$$ using its properties, identify the constants $$a$$, $$h$$, and $$k$$. Here, $$a = -2$$, $$h = 2$$, and $$k = -3$$. Since $$a = -2$$ is negative, the parabola opens leftward. The axis of symmetry is $$y = -3$$, and the vertex is $$(2, -3)$$. To find the $$x$$-intercept, set $$y = 0$$: $$x = -2(0+3)^2 + 2 = -2(9) + 2 = -16$$. The $$x$$-intercept is $$(-16, 0)$$. The symmetric point to this across the axis of symmetry $$y = -3$$ is $$(-16, -6)$$. Find the $$y$$-intercepts by setting $$x = 0$$: $$0 = -2(y+3)^2 + 2$$, which simplifies to $$1 = (y+3)^2$$. Solving this yields $$y+3 = \pm 1$$, giving $$y = -2$$ and $$y = -4$$. The $$y$$-intercepts are $$(0, -2)$$ and $$(0, -4)$$. These key features are then plotted to form the graph.

Graphing $$x = -2(y+3)^2 + 2$$

To graph the horizontal parabola $$x = 2y^2 + 12y + 17$$, first rewrite the equation in standard form, $$x = a(y - k)^2 + h$$, by completing the square. Factor $$2$$ from the $$y$$ terms: $$x = 2(y^2 + 6y) + 17$$. Complete the square inside the parentheses by adding $$9$$, and subtract $$18$$ (since $$2 \cdot 9 = 18$$) on the outside to keep the equation balanced: $$x = 2(y^2 + 6y + 9) + 17 - 18$$. This simplifies to standard form: $$x = 2(y + 3)^2 - 1$$. 

Now, identify the properties. The constants are $$a = 2$$, $$h = -1$$, and $$k = -3$$. Because $$a = 2$$, which is positive, the parabola opens to the right. The axis of symmetry is $$y = k$$, so $$y = -3$$. The vertex is $$(h, k)$$, which is $$(-1, -3)$$. Find the $$x$$-intercept by setting $$y = 0$$: $$x = 2(0 + 3)^2 - 1 = 2(9) - 1 = 17$$, giving the point $$(17, 0)$$. Find the symmetric point across the axis of symmetry, which is $$(17, -6)$$. Find the $$y$$-intercepts by setting $$x = 0$$: $$0 = 2(y + 3)^2 - 1$$, so $$\frac{1}{2} = (y + 3)^2$$. Solving gives $$y + 3 = \pm \frac{\sqrt{2}}{2}$$, or $$y = -3 \pm \frac{\sqrt{2}}{2}$$. This yields approximate $$y$$-intercepts at $$y \approx -2.3$$ and $$y \approx -3.7$$. Finally, plot these points and draw a smooth curve to graph the parabola.

Graphing $$x = 2y^2 + 12y + 17$$

To graph the horizontal parabola $$x = 3y^2 + 6y + 7$$, first rewrite the equation in standard form by completing the square. Factor out the $$3$$ from the $$y$$ terms: $$x = 3(y^2 + 2y) + 7$$. Complete the square inside the parentheses by adding $$1$$, and subtract $$3$$ outside to balance the equation (since $$3 \cdot 1 = 3$$): $$x = 3(y^2 + 2y + 1) + 7 - 3$$. This simplifies to the standard form: $$x = 3(y + 1)^2 + 4$$. 

From this standard form, $$x = a(y - k)^2 + h$$, identify the constants: $$a = 3$$, $$h = 4$$, and $$k = -1$$. Because $$a = 3$$, the parabola opens to the right. The axis of symmetry is the horizontal line $$y = -1$$, and the vertex is $$(4, -1)$$. Next, find the intercepts. Substitute $$y = 0$$ to find the $$x$$-intercept: $$x = 3(0 + 1)^2 + 4 = 7$$, giving the point $$(7, 0)$$. Its symmetric point across the axis of symmetry is $$(7, -2)$$. To find any $$y$$-intercepts, set $$x = 0$$: $$0 = 3(y + 1)^2 + 4$$, which leads to $$(y + 1)^2 = -\frac{4}{3}$$. Since the square of a real number cannot be negative, there are no real solutions, meaning the graph has no $$y$$-intercepts. Finally, plot the vertex, the $$x$$-intercept, and its symmetric point to draw the parabola.

Graphing $$x = 3y^2 + 6y + 7$$

To graph the horizontal parabola $$x = -4y^2 - 16y - 12$$, begin by writing the equation in standard form via completing the square. Factor $$-4$$ from the $$y$$ terms: $$x = -4(y^2 + 4y) - 12$$. Complete the square inside the parentheses by adding $$4$$. To balance the equation, subtract the equivalent value on the outside; since $$-4 \cdot 4 = -16$$, you add $$16$$ outside: $$x = -4(y^2 + 4y + 4) - 12 + 16$$. This simplifies to standard form: $$x = -4(y + 2)^2 + 4$$. 

Identify the properties from the standard form $$x = a(y - k)^2 + h$$: $$a = -4$$, $$h = 4$$, and $$k = -2$$. Since $$a = -4$$, the parabola opens to the left. The axis of symmetry is $$y = -2$$, and the vertex is $$(4, -2)$$. Find the $$x$$-intercept by setting $$y = 0$$: $$x = -4(0 + 2)^2 + 4 = -4(4) + 4 = -12$$, which gives the point $$(-12, 0)$$. Its symmetric point is $$(-12, -4)$$. Find the $$y$$-intercepts by setting $$x = 0$$: $$0 = -4(y + 2)^2 + 4$$, which gives $$(y + 2)^2 = 1$$. Solving this yields $$y + 2 = \pm 1$$, so $$y = -1$$ and $$y = -3$$. The $$y$$-intercepts are $$(0, -1)$$ and $$(0, -3)$$. Plot these points and draw a smooth curve to complete the graph.

Graphing $$x = -4y^2 - 16y - 12$$

A civil engineering technician is reviewing the design specifications for a horizontal parabolic support beam represented by the equation $x = a(y - k)^2 + h$. Match each property of the parabola with the correct description or formula used in the standard graphing procedure.

An industrial designer is mapping the profile of a parabolic reflector defined by the equation $x = ay^2 + by + c$. To ensure the profile is graphed accurately for a manufacturing blueprint, the designer follows a specific six-step procedure. Arrange these steps in the correct order as they should be performed.

A manufacturing technician is programming a CNC router to cut a horizontal parabolic groove defined by the equation $x = ay^2 + by + c$. To ensure the groove opens to the right on the metal plate, which condition must the technician recall regarding the coefficient $a$?

A civil engineering technician is mapping a parabolic drainage channel defined by the equation $x = ay^2 + by + c$. True or False: To correctly graph this horizontal parabola using the standard 6-step procedure, the technician must recall that the axis of symmetry is a horizontal line.

An industrial designer is following a six-step procedure to graph a horizontal parabola, $x = ay^2 + by + c$, for a structural component. After successfully finding the $x$-intercept in Step 4, what additional point must the designer locate according to this procedure before proceeding to the $y$-intercepts in Step 5?

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