How do you find the vertex and the intercepts for \[x = \dfrac{1}{{ - 32}}{y^2}\]?

Hint:
We can find the vertex and intercepts of the equation by their general formulas obtained and to find the y-intercept of the given equation, just we need to substitute \[x\]= 0 in the given equation and solve for y and to find the x-intercept of the given equation, just we need to substitute y = 0 in the given equation and solve for x.

Complete step by step solution:
The given equation is
\[x = \dfrac{1}{{ - 32}}{y^2}\].
The given equation is in the form of \[x = a{y^2} + by + c\], in which we need to find the vertex of x and y coordinate.
As the coefficient of \[{y^2}\]is negative, considering equation \[y = a{x^2} + bx + c\] the b from bx moves the graph left or right in that
\[{x_{vertex}} = \left( { - \dfrac{1}{2}} \right) \times b\]
Now considering equation \[x = a{y^2} + by + c\] the b from by moves the graph up or down in that
\[{y_{vertex}} = \left( { - \dfrac{1}{2}} \right) \times b\]
However, we get b=0, so the axis of symmetry coincides with the x-axis.
Hence, y-intercept is at \[x = 0\].
Substitute the value of x in the given equation as
\[x = \dfrac{1}{{ - 32}}{y^2}\]
\[0 = \dfrac{1}{{ - 32}}{y^2}\]
\[ \Rightarrow y = 0\]
Therefore, as there is only a single value as it does not cross the y-axis but y-axis is tangential to the vertex.
Hence, the vertex of x and y is \[\left( {0,0} \right)\].
The intercepts are only at one point i.e., at the origin and the axis of symmetry is the x-axis i.e., y = 0.

Note:
As per the given equation consists of x and y terms based on the intercept asked, we need to solve for it. For ex if y-intercept is asked substitute x=0 and solve for y and if x-intercept is asked substitute y=0 and solve for x and the y-intercept of an equation is a point where the graph of the equation intersects the y-axis.