In order to determine which direction the parabola opens, we must first put the equation in standard form, which can be expressed in one of the following two ways:

$\displaystyle y=ax^2+bx+c$

$\displaystyle x=ay^2+by+c$

If the equation is for $\displaystyle y$ as in the first above, the parabola opens up if $\displaystyle a$ is positive and down if $\displaystyle a$ is negative. If the equation is for $\displaystyle x$ as in the second above, the parabola opens right if $\displaystyle a$ is positive and left if $\displaystyle a$ is negative. Rearranging our equation, we get:

$\displaystyle x=-y^2+6y+13$

We can see that our equation is for $\displaystyle x$, which means the parabola will open either left or right. The sign of the first term is negative, so this parabola will open to the left.

For the function

$\displaystyle y=ax^2$

The parabola opens upwards if a>0

and downards for a<0

Because 

$\displaystyle 3>0$

The parabola opens upwards.

The standard form for a parabola is in the form: $\displaystyle ax^2+bx+c$

The coefficient of the $\displaystyle a$ term determines whether if the parabola opens upward or downward.  Since the $\displaystyle a$ term in the function $\displaystyle y=-5x^2+3x-4$ is $\displaystyle a=-5$, the parabola will open downward.

Use the FOIL method to determine $\displaystyle y=(1-x)(x-2)$ in its standard form for parabolas, which is $\displaystyle y=ax^2+bx+c$.  

$\displaystyle y=(1-x)(x-2)$

$\displaystyle y= (1)(x)+(1)(-2)+(-x)(x)+(-x)(-2)$

$\displaystyle y= x-2-x^2+2x$

Regroup the terms.

$\displaystyle y=-x^2+3x-2$

Since the coefficient of the $\displaystyle a$ term is negative, the parabola will open downward.

The focus is above the vertex, which means that the parabola will open up

In order to determine which way this parabola, group the variables in one side of the equation.  Add $\displaystyle y$ on both sides of the equation to isolate $\displaystyle x$.

$\displaystyle x-y=-2y^2-2$

$\displaystyle x=-2y^2+y-2$

Because the equation is in terms of $\displaystyle y$, the parabola will either open left or right.  Notice that the coefficient of the $\displaystyle y^2$term is negative.

The parabola will open to the left.

Determine whether the following parabola opens up or down and state how you know.

$\displaystyle y=3x^2-4x-198$

To determine the direction a parabola opens, we only need to worry about the squared term. 

In this case, it is positive, so the parabola opens upward.

The linear term is negative, so the parabola will be to the right of the y-axis.

The constant term is negative, so the parabola will be located below the x-axis.

In order to determine how the parabola will open, we will need to rewrite the equation in standard form.  

Write the standard form for parabolas.

$\displaystyle y=ax^2+bx+c$

Subtract $\displaystyle -2x^2$ from both sides of the equation.

$\displaystyle 2x^2+y-2x^2=-2x^2+4$

$\displaystyle y=-2x^2+4$

Since the coefficient of the $\displaystyle a$ is negative, and the equation is in terms of $\displaystyle x$, the parabola will open downward.

The answer is:  $\displaystyle \textup{Down}$

The answer is $\displaystyle y = -(x+1)^2 - 3$ because it is the only degree-2 polynomial with a negative leading coefficient.

In order to be in standard form, the equation for a parabola must be written in one of the following ways:

$\displaystyle y=ax^2+bx+c$

$\displaystyle x=ay^2+by+c$

We can see that our equation has all of the components of the first form above, so now all we must do is some algebra to rearrange the equation and express the function y in terms of x. We start by simplifying the fraction on the left side of the equation, and then we isolate y to give us the equation of the parabola in standard form:

$\displaystyle \frac{x^2-6x}{3}=y+7$

$\displaystyle \frac{1}{3}x^2-2x=y+7$

$\displaystyle y=\frac{1}{3}x^2-2x-7$