                   A quadratic function has the form f(x) = ax2 + bx + c , where a, b, and c are real number constants and a ¹ 0.  Its domain is all real numbers.  The constants a, b, and c are called the coefficients of the quadratic function. The graph of a quadratic function is a parabola opening up if the leading coefficient a > 0 or opening down if a < 0.

The simplest quadratic functions are y = x2 and y = -x2. The graphs are shown below.  y = x2                                                                                       y = -x2

To see how different values of the coefficients affect the graph, experiment with the following interactive example.

### Example

The function H(t) = -16t2  + 45t + 200 describes the height in feet of a ball t seconds after it is thrown upward from the top of a 200 foot high building. The graph below shows that the ball hits the ground shortly after 5 seconds and reaches its maximum height of approximately 230 feet after about 1.5 seconds. We will learn to calculate these exact values later. ## Intercepts

The graph of a quadratic function will always have one y-intercept and can have as many as two x-intercepts (recall that intercepts are points where a graph crosses an axis).  The y-intercept is the point where the graph crosses the y-axis so it will be the value of the function at x = 0.  Since f(0) = c, the y-intercept is the point (0,c).  The x-intercept(s), if there are any, are the points where the graph crosses the x-axis.  They can be found by replacing y with 0 and solving the resulting quadratic equation for x. This equation has one, two, or no solutions.  When it has one solution, there will be one x-intercept; when there are two solutions, there will be two x-intercepts and when the equation has no solution, the graph will have no x-intercepts.  To see examples, expolore the interactive example below.

### Example

Find the x- and y- intercepts of the quadratic function

f(x) = 1.5x2 - 3.2x - 9.6.

### Solution

The y-intercept is found by replacing x with 0:

f(0) = 1.5(0)2 - 3.2(0) - 9.6 = -9.6.

Thus the y-intercept is the point (0, -9.6).

The x-intercepts are found by replacing y or f(x) with 0 and solving for x:

0 = 1.5x2 - 3.2x - 9.6.

Using the quadratic formula, we get The solutions (rounded to two decimal places) are x = 3.81 and x = -1.68.  The x-intercepts are approximately the points (3.81, 0) and (-1.68, 0).  A graph is shown below. TRY THESE PROBLEMS NOW

Decide whether each parabola opens up or down and determine the coordinates of the x- and y-intercepts.

1.         y = .02x2 - 63.1x + 27

2.         P(t) = -3t2 + 27.5t + 13.1

## Vertex

A parabola which opens up has a lowest point and a parabola which opens down has a highest point.  The highest or lowest point on a parabola is called the vertex.  The parabola is symmetric about a vertical line through its vertex, called the axis of symmetry. The figure below shows a parabola opening up with vertex (0.75, 0.875) and axis of symmetry x = 0.75. ### Vertex Form of a Quadratic Function

To find the vertex of a parabola, we will write the function in the form .  As an example, consider the function . We first complete the square on the right side:

f(x) = 2(x2 - 4x) + 7                  (factor out 2 from the terms 2x2 - 8x)

= 2(x2 - 4x + 4) + 7 - 8       (complete the square of x2 - 4x)

= 2(x - 2)2 - 1                                 (factor the perfect square and simplify.)

Notice that for all values of x.  Thus f(x) = 2(x - 2)2 - 1 ³ -1 for all values of x and the minimum value of the function is -1 when x = 2.  The point (2, -1) is the lowest point on the graph so it is the vertex of the parabola. The vertical line x = 2 is the axis of symmetry.  See the graph below.  In general, is called vertex form of a quadratic function.  When a quadratic function is written in vertex form, we can easily determine the vertex (h, k). If the coefficient a > 0, then the parabola opens upward and the vertex is the lowest point on the parabola. We say that k is the minimum value of the quadratic function. On the other hand, if the coefficient a < 0, then the parabola opens downward and the vertex is the highest point on the parabola. In this case, k is the maximum value of the quadratic function. Explore the role of each coefficient in the following interactive example.

### Example

Write the quadratic function in vertex form. Determine the vertex and the maximum or minimum value of the function.

### Solution

We will complete the square to write the function in vertex form: The vertex form is , so the vertex is (3, -11). Since a < 0, the parabola opens downward and the vertex is the highest point. The function has a maximum value of 11. Its graph is shown below.  Once we know the vertex of a parabola, we can determine the range of the quadratic function. Consider the function, . Previously we determined that the parabola has a minimum value of -1, occurring when x = 2. Thus the range of the quadratic function is {y½ y ³ -1}. As another example, lets return to the function in the above example. The graph of this function is a parabola opening downward and the maximum value of the function is 11. Therefore, the range of the quadratic function is ### Finding the Vertex Algebraically

The vertex of a quadratic function can also be determined algebraically. We first assume that the quadratic function has two x-intercepts. Then the graph is a parabola that crosses the x-axis in two distinct points. Since the parabola is symmetric with respect to a vertical line through its vertex (the axis of symmetry) the x-coordinate of the vertex is always halfway between the two x-intercepts. By the quadratic formula, the two x-intercepts are Notice that the same number, , is being added to and subtracted from . It follows that the number is halfway between This means that the x-coordinate of the vertex is . We can then find the y-coordinate of the vertex by evaluating Although we assumed that the quadratic function had two x-interecpts when we derived our vertex formula, it also holds in the other two cases, where the parabola has one or no x-intercepts.

### Example

Find the vertex of the quadratic function . Use the vertex to determine the maximum or minimum value of the function and find its range.

### Solution

The vertex formula gives To find the second coordinate of the vertex, we evaluate The vertex of the parabola is (3, 53). Since a < 0, the parabola opens downward and the vertex is the highest point. This gives a maximum value of 53 and the range of the function is A graph of the function is shown below.  Recall the function , which describes the height in feet of a ball t seconds after it is thrown upward from the top of a 200 foot high building. We can now determine when the ball hits the ground and the maximum height that it reaches, as well as the time that it reaches that maximum height. When the ball hits the ground, its height above ground will be zero. This gives the quadratic equation . Using the quadratic formula, we find that the solutions are and (rounded to two decimal places). Since the time cannot be negative, we see that the ball strikes the ground after 5.21 seconds. The maximum height of the ball will be given by the second coordinate of the vertex and the time will be the first coordinate. Using the vertex formula we find that (rounded  to tow decimal places). Next we evaluate This means that the ball reaches its maximum height of 231.64 feet after 1.41 seconds.  