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:
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:
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
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
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.
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: f(x) + c causes the graph to undergo a vertical shift upward c units if c > 0; f(x) + c causes the graph to undergo a vertical shift downward c units if c < 0.
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
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).
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 right of left--depending on whether c is positive or negative; If y = f(x) and we consider f(x) + c, the parabola shifts c units up or down--depending on the sign of c. If the leading coefficient a > 0, the parabola opens up (or is concave up); if a < 0, the parabola opens down.