Graphing Functions: Completing the Square, Quadratic Functions, Axis of Symmetry, and Intercepts

1.

This parabola has a vertex at

We see that as the function approaches , so the vertex point is a maxima (the parabola opens down).
The parabola intersects the x-axis when

 
Which gives:

It intersects the y-axis when x=0 and this gives:

To summarize:
The parabola has a vertex which is a maxima at so its symmetry axis is x=1
It intersects the x-axis at and
It intersects the y-axis at
 
It looks like:

 
2.
The function is

This is a third degree polynomial that has one root with multiplicity 3.
When we have
The graph is the same as that of only transposed 2 units to the right.
It intersects the y-axis when x=0 so the intersection point is
It looks like:

 
3.
The general form of a quadratic function is:

Where is the vertex.
In our case the vertex is , hence our parabola has the form:

We require that , and this gives for a:

Therefore:

 
Opening the parenthesis

It looks like:

 
4.
A line with slope m and y-intercept b has the canonical equation:

In our case m=0, so the line's equation is:

This is an equation of a line parallel to the x axis (y is constant)
Since the line passes through the point (2,-7) we see this constant must be -7, hence the line equation is

And it looks like:

5.
The function is:

We convert it to the canonical form :

We see that the vertex is

And it is a minima (the parabola opens upward)
The function is symmetric about the vertical axis that passes through the vertex:

It intersects the x axis when which gives:

The x-intercepts are and
It intersects the y-axis when x=0 which gives:

The y-intercept is

It looks like:

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!