Notes, Lesson 3.1

One way to solve a quadratic function is to think of it as a transformation of the basic quadratic function: . We will be striving to get our quadratic equations in the form: . By putting our quadratics in this form, we can then think of our quadratics as transformations of . Our quadratics will then show whether they open up or down (by looking at the sign of a, If they are standard width or narrower or wider (by looking at the size of a), whether the graph is shifted to the left or right and how much (by looking at the value of h), and whether the graph is shifted up or down (by looking at the value of k).

To achieve this form, we will use the completing the square technique. This completing of the square will be slightly different than the technique used previously. Let's look at an example: Write this equation in the form and sketch its graph. We first factor out the -2 coefficient of x squared only from the two x terms. Now, we complete the square on the quantity in parentheses. We took half of the coefficient of x and squared it. Then we added this term (1) in the parentheses. Because we really just subtracted 2 from the equation (because the entire quantity is multiplied by -2), we must also add 2 so that this is a legal operation. We have now gotten to our form. Now we can see that 1)the graph open downward (because of the negative a term, 2) the graph is thinner than usual (because a is > 1, 3) From the origin it is moved to the left 1 (because of h = 1), 4) it is moved up 10 units (because k = 10). Here we see the graph of this function. Does it confirm our findings in the previous step? It does. Please notice that the graph does not appear to be narrower, but if you look at the scale of the y-axis, you can see that it is indeed narrower.

Another example: Find 1) the range; 2) the maximum or minimum value of the function; 3) identify the intervals where the function is increasing or decreasing. Again, we want to use the completing the square technique to find out more about the function. We begin by factoring out the -3 out of the x terms only. Now we ready to complete the square on the quantity in parentheses. We take have of the coefficient of x (3 in this case) and we square the result. This 9/4 then is to be added in the parentheses to make the quantity in the parentheses a perfect square trinomial. Because we really have just subtracted 27/4 from the equation, we must equalize things by adding the same 27/4 outside of the parentheses. Now we write the perfect square trinomial as a binomial squared, and simplify the numbers outside the parentheses. We now have the information needed to answer the questions. We know that this parabola opens downward (negative a value). This tells us that there is a maximum, but no minimum. The maximum will be at the y value at the vertex. On what interval for x is the function increasing? Before the vertex. On what interval for x is the function decreasing? After the vertex.

Finding Intercepts. One important feature of a quadratic function, is the ability to find the x and y intercept(s). To find the y-intercept, set x = 0 and solve for y. This is usually the easy one to find. To find the x-intercepts, set y = 0 and solve for x. We have done this before. When a quadratic equation is set equal to zero, we know how to:

1) solve by completing the square;

2) solve by factoring;

3) solve by using the quadratic formula.

Any of these techniques can be used to find these x-intercepts. These x-intercepts are also called solutions, zeros, or roots of the equation.

Example Problem: Identify 1) the vertex, 2) the axis of symmetry, 3) the y-intercepts, 4) the x-intercepts, and 5) the opening of the parabola, and then graph the function. One of the beauties of this form of a quadratic function, is that you can immediately see the vertex. We know that it is a standard quadratic function moved to the left one unit, and down 9 units. Therefore we know the vertex. Another very obvious result is the equation of the axis of symmetry. The axis of symmetry must go through the vertex. This is the easy intercept to find. Substitute zero for x and solve for y. To find the x-intercepts, we must set y = 0 and solve for x. Before we do that, we need to get our equation in its multiplied out format. We could use any of our 3 methods for solving this equation. (listed above). Factoring works nicely here. Setting each factor equal to zero and solving gives us these two solutions. This is because in this problem a is positive. Finally, we graph the function and check our results above.

Quadratic Inequalities. We need our skills above and some common sense to solve quadratic inequalities. We again strive to get zero on one side of the inequality and then instead of asking the question where does the parabola cross the x-axis, we now want to answer a question like this: where is the parabola above (or below) the x-axis?

Sample Problem: Solve the inequality. It is still important to know where the points of intersection with the x-axis are. (if there are any). So we still use our old skills of finding the roots of a quadratic equation. In this case again, factoring works very easily. In fact, we have a perfect square trinomial here. For each of the factors, you would set them equal to zero and you would get this same result both times. What does that mean that both x-intercepts are the same? ... that the parabola only has 1 crossing point on the x-axis. Because the parabola opens upward, that means the entire parabola is above the x-axis. Now back to the original question. when is this parabola on or above the x-axis?  Answer: always. In interval notation we would write:  The graph confirms our analysis above. We can see that the parabola touches the x-axis at -1/3, and that the entire parabola is above (or on) the x-axis.