Recall the standard form of the equation of a vertical parabola:

\(\displaystyle (x-h)^2=4p(y-k)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) gives the focal length.

When \(\displaystyle p>0\), the parabola will open up.

When \(\displaystyle p< 0\), the parabola will open down.

For the parabola in question, the vertex is \(\displaystyle (-2, 0)\) and \(\displaystyle p=-3\). This parabola will open down. Because the parabola will open down, the focus will be located \(\displaystyle 3\) units down from the vertex. The focus is then located at \(\displaystyle (-2, -3)\)

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

For the given parabola, the vertex is \(\displaystyle (0, 3)\) and \(\displaystyle p=-1\). This means the parabola is opening to the left and that the directerix will be located \(\displaystyle 1\) unit to the right of the vertex. The directerix is then \(\displaystyle x=1\).

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

For the given parabola, the vertex is \(\displaystyle (-1, 4)\) and \(\displaystyle p=-2\). This means the parabola is opening to the left and that the directerix will be located \(\displaystyle 2\) units to the right of the vertex. The directerix is then \(\displaystyle x=1\).

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

Put the given equation into the standard form. Start by isolating the \(\displaystyle y\) terms to one side.

\(\displaystyle y^2+2y-12x+13=0\)

\(\displaystyle y^2+2y=12x-13\)

Complete the square. Remember to add the same amount to both sides of the equation.

\(\displaystyle y^2+2y+1=12x-13+1\)

\(\displaystyle y^2+2y+1=12x-12\)

Factor both sides of the equation to get the standard form of the equation of a horizontal parabola.

\(\displaystyle (y+1)^2=12(x-1)\)

For the given parabola, the vertex is \(\displaystyle (1, -1)\) and \(\displaystyle p=3\). This means the parabola is opening to the right and that the directerix will be located \(\displaystyle 3\) units to the left  of the vertex. The directerix is then \(\displaystyle x=-2\).

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

For the given parabola, the vertex is \(\displaystyle (0, -1)\) and \(\displaystyle p=4\). This means the parabola is opening to the right and that the focus will be located \(\displaystyle 4\) units to the right of the vertex. The focus is then located at \(\displaystyle (4, -1)\).

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

For the given parabola, the vertex is \(\displaystyle (7, 5)\) and \(\displaystyle p=5\). This means the parabola is opening to the right and that the focus will be located \(\displaystyle 5\) units to the right of the vertex. The focus is then located at \(\displaystyle (12, 5)\).

Recall the standard form of the equation of a horizontal parabola:

\(\displaystyle (y-k)^2=4p(x-h)\), where \(\displaystyle (h, k)\) is the vertex of the parabola and \(\displaystyle |p|\) is the focal length.

When \(\displaystyle p>0\), the parabola opens to the right.

When \(\displaystyle p< 0\), the parabola opens to the left.

Start by putting the equation into the standard form of the equation of a horizontal parabola. 

Isolate the \(\displaystyle y\) terms on one side.

\(\displaystyle y^2+12y-24x+60=0\)

\(\displaystyle y^2+12y=24x-60\)

Complete the square. Remember to add the same amount to both sides of the equation!

\(\displaystyle y^2+12y+36=24x-60+36\)

\(\displaystyle y^2+12y+36=24x-24\)

Factor both sides of the equation to get the equation in the standard form.

\(\displaystyle (y+6)^2=24(x-1)\)

For the given parabola, the vertex is \(\displaystyle (1, -6)\) and \(\displaystyle p=6\). This means the parabola is opening to the right and that the focus will be located \(\displaystyle 5\) units to the right of the vertex. The focus is then located at \(\displaystyle (7, -6)\).

We can first re-write the equation by multiplying both sides by 4:

\(\displaystyle 4y = (x-7)^2\)

This equation tells us that the vertex for the parabola is \(\displaystyle (7, 0)\)

We can also find the distance from the vertex to the focus and the directrix by setting up the equation \(\displaystyle 4 = 4p\). This gives us \(\displaystyle 1 = p\), so the focus and the directrix are both 1 away from the vertex.

Since this equation is positive, the parabola opens up. This means the directrix is one below the vertex, so it is at \(\displaystyle y = -1\). The focus is one above the vertex, so it is at \(\displaystyle (7,1)\).

Before we can start determining the focus and directrix, we need to re-write this equation in standard form. That way we can see the vertex and we can more easily determine the distance from the vertex and the focus/directrix.

To do this, we should complete the square. First subtract 10 from both sides:

\(\displaystyle y - 10 = x^2 + 4 x\)

Now we need to figure out what to add to both sides in order to make the right side a perfect square. We should add 4, since half of the 4 in 4x is 2, and \(\displaystyle 2^2 = 4\)

\(\displaystyle y - 10 + 4 = x^2 + 4 x + 4\)

We can simplify the left side and re-write the right side as a binomial squared: 

\(\displaystyle y - 6 = (x+2)^2\)

Interpreting this, we determine that the vertex is \(\displaystyle (-2, 6)\). On the left side, \(\displaystyle y-6\) is multiplied by 1, so \(\displaystyle 4p = 1\) where p measures the distance from the vertex to the focus/directrix. Solving this gives us \(\displaystyle p = \frac{1}{4}= 0.25\). 

Since this is a positive, upward-opening parabola, the directrix is below the vertex and the focus is above. The focus is then located at \(\displaystyle (-2, 6+0.25 ) = (-2, 6.25)\). The directrix is then located at \(\displaystyle y = 6 - 0.25 = 5.75\).

For the function

\(\displaystyle y=ax^2\)

The parabola opens upwards if a>0

and downards for a<0

Because 

\(\displaystyle -5< 0\)

The parabola opens downwards.