In this assignment we are examining the quadratic function

and its graph. Let's begin by actually looking at its graph.

We may make several observations at this point. First, notice that the
parabola opens up. It follows that the vertex of this parabola is lowest
point on the graph. We may find the vertex of any parabola as follows:

where a, b, and c are real numbers,

the x-coordinate of the vertex may be calculated by

.

Of course, we way find the y-coordinate of the vertex by simply substituting
the value of x back into the original equation. In order to calculate the
vertex of this equation, we know that

Since we now know the x-coordinate of the vertex, we may find the y-coordinate
by

At this point, we may express the vertex as the ordered pair (-.75, -5.125).
Of course, we should also been able to detect this from our graph just by
"eyeballing" and getting an approximate location.

We should also note that the parabola has two x-intercepts. This means that
if we were to solve this as a quadratic equation, we would have two real
roots. Although the equation will not factor conventionally, we may use
the quadratic formula:

where

,

and a, b, and c are real, a is different from 0.

Using the quadratic formula we soon discover that our solutions are -2.351
and .851. If we choose to express them as solutions we should write them
as a set such as {-2.351, .851}. On the other hand, if we choose to express
our answers as the x-intercepts, they should be listed as the points (-2.351,
0) and (.851, 0).

In addition to x-intercepts, we should also mention the y-intercept. It
is not difficult to see from the graph that our function crosses the y-axis
at -4, or at the point (0, -4). If we wish to find the y-intercept algebraically,
we may simply find f(0) if

We have already pointed out many of the attributes of the function

Now, let's evaluate f(x - 4) and make note of how the attributes change.
We shall begin by comparing the two graphs.

The red graph represents the original function, whereas the green graph
represents the function

Notice that the entire graph has simply shifted 4 units to the right
of its original location. This implies we are simply taking each of the
x-coordinates and adding 4 units while leaving the y-coordinates unchanged.
Before we try to generalize this situation, it would be wise to investigate
some other graphs. Consider the functions

and their graphs.

Again, our original graph is displayed in red, f(x - 5) is displayed
in green. It is important to make mention of the fact that the size and
shape of the graph did not change, the graph has simply shifted 5 units
to the right. The functions f(x + 1) (displayed in blue) and f(x + 3) (displayed
in gold) simply shifted the graph 1 unit to the left and 3 units to the
left, respectively.

We may now generalize. Consider the quadratic function

where c is some real number. If c > 0, then the graph undergoes a
horizontal shift c units to the right. If c < 0, then the graph undergoes
a horizontal shift c units to the left.

We have already seen our function undergo horizontal shifts, but what causes the function to shift vertically. Certainly, we have seen that altering x changes the graphs vertical position, let's try changing y. Let's begin by graphing f(x) + 7. Since

our new function is

Let's compare the two graphs.

Notice that our graph has been shifted 7 units vertically. Moreover,
we no longer have x-intercepts so this implied that our solutions to the
quadratic equation would be imaginary. The y-intercept has also changed
from (0, -4) on the original graph to (0, 3) on the green graph. It follows
that we are adding 7 units to each of the y-coordinates on the original
graph. Finally, by graphing f(x) + 7, we have also changed the location
of the vertex from the third quadrant to the second quadrant--which is one
of the goals of this exploration.

Let's generalize the above findings:

The last goal of this investigation is to produce a graph that is concave down, yet shares the same vertex as the quadratic function

Let's try to graph the original function as well as the function

and make the necessary comparisons. Before we attempt to graph the functions,
it is good practice to transform each equation from quadratic form into
standard form, which is

This procedure requires completing the square. Our two equations in standard
form are

and

.

Now, let's graph these two functions.

Notice that the two parabolas are tangent at the vertex. Once the two
functions have been transformed into standard form, we may simply negate
the leading coefficient and this causes the graph to reflect over the line
y = -5.125. We should also note that by reflecting the graph, there are
no x-intercepts and both solutions to this quadratic equation will be imaginary.
It is rather interesting that the leading coefficient is same regardless
of whether the equation is expressed in quadratic form or whether it is
expressed in standard form. It should also be pointed out that the vertex
may be ascertained directly from the equation if the it is in standard form.
Indeed, the vertex is given as (h , k).

or

where a, b, and c are real numbers, a is different than 0, is considered
a quadratic and its graph is a parabola.

If y = f(x) and we consider f(x) + c, the parabola shifts c units up or
down--depending on the sign of c.