Equations of Simple Parabolas
You may want to review the prior web exercise, Parabolas.
If a parabola is placed in a coordinate plane in a simple way, then a simple equation is obtained.
 Derivation: parabola with vertex at origin, focus on $y$axis
 Equations of simple parabolas
 Memory Device: ‘one over four pee pairs’
 Shifting the parabola
 Graphing a shifted parabola
Derivation
Place a parabola with its vertex at the origin, as shown below.

If we put the focus on the $y$axis, then the directrix will be parallel to the $x$axis.

Or, if we put the directrix parallel to the $x$axis, then the focus will be on the $y$axis.
In either case, let $\,p \ne 0\,$ denote the $y$value of the focus. Thus, the focus has coordinates $\,(0,p)\,.$
Although the sketch above shows the situation where $\,p\gt 0\,,$ the following derivation also holds for $\,p \lt 0\,.$
Notice that:
$p\gt 0$  if and only if  the focus is above the vertex  if and only if  the parabola is concave up (holds water) 
AND  
$p\lt 0$  if and only if  the focus is below the vertex  if and only if  the parabola is concave down (sheds water) 
For example, if the focus is above the vertex, then $\,p\,$ must be greater than zero, and the parabola must be concave up. As a second example, if the parabola is concave down, then $\,p\,$ must be less than zero, and the focus is below the vertex.
Since the vertex is a point on the parabola, the definition of parabola dictates that it must be the same distance from the focus and the directrix.
Thus, the directrix must cross the $y$axis at $\,p\,$; indeed, every $y$value on the directrix equals $\,p\,.$
Let $\,(x,y)\,$ denote a typical point on the parabola.
The distance from $\,(x,y)\,$ to the focus $\,(0,p)\,$ is found using the distance formula: $$ \tag{1} \cssId{s36}{\sqrt{(x0)^2 + (yp)^2 } = \sqrt{x^2 + (yp)^2}} $$
To find the distance from $\,(x,y)\,$ to the directrix, first drop a perpendicular from $\,(x,y)\,$ to the directrix. This perpendicular intersects the directrix at $\,(x,p)\,.$ The distance from $\,(x,y)\,$ to the directrix is therefore the distance from $\,(x,y)\,$ to $\,(x,p)\,$:
$$ \tag{2} \cssId{s40}{\sqrt{(xx)^2 + ((y(p))^2} = \sqrt{(y+p)^2}} $$From the definition of parabola, distances $\,(1)\,$ and $\,(2)\,$ must be equal:
$$ \cssId{s42}{\sqrt{x^2 + (yp)^2} = \sqrt{(y+p)^2}} $$This equation simplifies considerably, as follows:
Squaring both sides:  $ x^2 + (yp)^2 = (y+p)^2 $ 
Multiplying out:  $ x^2 + y^2  2py + p^2 = y^2 + 2py + p^2 $ 
Subtracting $\,y^2 + p^2\,$ from both sides:  $ x^2  2py = 2py $ 
Adding $\,2py\,$ to both sides:  $ x^2 = 4py $ 
Dividing by $\,4p\,$ and rearranging:  $ \displaystyle y = \frac{1}{4p} x^2 $ 
Such a beautiful, simple description for our parabola! The most critical thing to notice is the coefficient of $\,x^2\,,$ since it holds the key to locating the focus of the parabola.
As an example, consider the equation $\,y = 5x^2\,.$ Comparing $\,y = 5x^2\,$ with $\,y = \frac1{4p}x^2\,,$ we see that $\, 5 = \frac{1}{4p}\,.$ Solving for $\,p\,$ gives:
$\displaystyle 5 = \frac{1}{4p}$  original equation 
$20p = 1$  multiply both sides by $\,4p$ 
$\displaystyle p = \frac{1}{20}$  divide both sides by $\,20$ 
Thus, $\,y = 5x^2\,$ graphs as a parabola with vertex at the origin, and focus $\,(0,\frac{1}{20})\,.$ So easy!
In summary, we have:
Every equation of the form $\,y= ax^2\,$ (for $\,a\ne0\,$) is a parabola with vertex at the origin, directrix parallel to the $x$axis, and focus on the $y$axis.
If $\,a\gt 0\,,$ then the parabola is concave up (holds water).
If $\,a\lt 0\,,$ then the parabola is concave down (sheds water).
If $\,p\,$ denotes the $y$value of the focus, then $\,a = \frac{1}{4p}\,.$ Solving for $\,p\,$ gives $\, p = \frac{1}{4a}\,,$ and thus the coordinates of the focus are $\,(0,\frac{1}{4a})\,.$
Memory Device: ‘one over four pee pairs’
Notice that if $\,a = \frac{1}{4p}\,,$ then $\,p = \frac{1}{4a}\,.$ Or, if $\,p = \frac{1}{4a}\,,$ then $\,a = \frac{1}{4p}\,.$ What a beautiful symmetric relationship!
Thus, I fondly refer to $\,a\,$ and $\,p\,$ with the catchy phrase: ‘one over four pee pairs’. (Try to say this quickly ten times in a row!)
If you know either $\,a\,$ or $\,p\,,$ then it's easy to find the other—just multiply by $\,4\,$ and then flip (take the reciprocal).
For example, if you know that $\,a = 5\,,$ then $\,p\,$ is found as follows:

$4\times 5 = 20\,$ (multiply by $\,4\,$)

$p=\frac{1}{20}$ (flip)
Or, if you know that $\,p = \frac{1}{20}\,,$ then $\,a\,$ is found as follows:

$4\times \frac{1}{20} = \frac{4}{20} = \frac{1}{5}\,$ (multiply by $\,4\,$)

$a = (\text{the reciprocal of } \frac{1}{5}) = 5\,$ (flip)
ReadThrough, Part 2
Shifting the Parabola
Now, here's some very good news. By using graphical transformations, knowledge of this one simple equation $\,y = ax^2\,$ actually gives full understanding of all parabolas with directrix parallel to the $x$axis! The results are summarized next.
Shift the parabola (together with its focus and directrix) horizontally by $\,h\,,$ and vertically by $\,k\,.$ This yields the following information:
Example (Graphing a Shifted Parabola)
In this example, I illustrate the approach that I usually take when asked to give complete information about the parabola $\,y = a(xh)^2 + k\,.$

Concave up or down? Since $\,a=3\lt 0\,,$ the parabola is concave down (sheds water). Since the focus is always inside a parabola, we know the focus is under the vertex.

Find the vertex: The vertex of $\,y = a(xh)^2 + k\,$ is the point $\,(h,k)\,$:
$$ \cssId{sb34}{y = a( \overset{\text{What value of } \ x\ }{\overset{\text{ makes this zero?}}{\overbrace{ \underset{\text{answer: } \ h}{\underbrace{xh}}}}})^2\ \ \ + \overset{y\text{value of the vertex}}{\overbrace{\ \ k\ \ }}} $$For us, what makes $\,x+5\,$ equal to zero? Answer: $\,5\,$
Thus, $\,5\,$ is the $x$value of the vertex. When $\,x = 5\,,$ the corresponding $y$value is $\,1\,.$ Thus, the vertex is $\,(5,1)\,.$

Find the distance from the vertex to the focus: Using the ‘one over four pee pair’ memory device:
$$\cssId{sb42}{p = \frac{1}{4a} = \frac{1}{4(3)} = \frac{1}{12}}$$Thus, the distance from the focus to the vertex is:
$$\cssId{sb44}{p = \left\frac1{12}\right = \frac{1}{12}}$$ 
Find the focus: We already determined that the focus lies below the vertex. So, the focus is:
$$\cssId{sb48}{(5,1\frac{1}{12}) =(5,\frac{11}{12})}$$ 
Find the equation of the directrix: Every horizontal line has an equation of the form:
$$\cssId{sb51}{y = \text{(some #)}}$$In our example, the focus lies $\,\frac{1}{12}\,$ below the vertex, so the directrix lies $\,\frac{1}{12}\,$ above the vertex. Thus, the equation of the directrix is $\,y=1+\frac{1}{12}\,,$ that is, $\,y=\frac{13}{12}\,.$

Plot an additional point: Plotting an additional point gives a sense of the ‘width’ of the parabola. For example, when $\,x = 4\,$ we have:
$$ \begin{align} y &= 3(4+5)^2 + 1\cr &= 3(1) + 1\cr &= 2 \end{align} $$In this parabola, the vertex is close to the focus, so the parabola is narrow.
$\,y=3(x+5)^2+1\,$
For fun, zip up to WolframAlpha and type in any of the following:
vertex of y = 3(x+5)^2 + 1
focus of y = 3(x+5)^2 + 1
directrix of y = 3(x+5)^2 + 1
How easy is that?!