Exercise 24 Page 607

Let's start by reviewing the simplest quadratic function, y=x^2. Its graph is a parabola with its vertex at the origin, this being its lowest point.

We can transform this graph if we multiply it by a constant factor a. y=x^2 → y = ax^2 If a< 0, all the function's values become negative and the graph is reflected across the x-axis. Then the parabola would be upside down. In this case, its vertex is the highest point and the parabola has a maximum value.

However we still have just one x-intercept. We can solve this by adding a constant value to the function. y = ax^2 → y = ax^2+c If c > 0, all the function's values are increased by the same constant amount and the graph is translated vertically upwards. Consequently, the parabola now intercepts the x-axis twice, as required. For example, consider the function y = - 12x^2 + 2.

As we can see, this function has two x-intercepts and a maximum value, just as the exercise requests. Notice that this is only an example, since any function of the form y= ax^2+c, with a<0 and c>0, would satisfy the exercise's requirements. Therefore, there are infinitely many solutions.