Quadratic Functions

 

A quadratic function always has an x2 term and is represented in graphical form by a parabola. The function has two forms, unlike the linear function that has five:

 

  1. y = ax2  + bx + c, where a ≠ 0, called the standard form,

 

  1. y = a(x – h)2 + k, where a ≠ 0, called the vertex form with (h, k) the vertex of the parabola.

 

In both forms, a, b, c, h, k are real numbers. Ask yourself, why is it important for a ≠ 0 to remain true?

 

As you can see in both forms of the quadratic function, a is the coefficient of the x2 term. The value of a gives us vital information about the function and its graph, the parabola.

 

  1. a > 0 (positive) tells us that the parabola direction opens upward. Obviously, the parabola opens downward when a is negative.

 

  1. a = 1 is the “standard” width parabola. Graph y = x2 to see the standard parabola.

 

  1. a >1 is a parabola more narrow than the standard parabola. Try this on your calculator.

 

  1. 0 < a < 1 is a parabola wider than the standard parabola. Try this on your calculator.

 

 

 

The parabola is a graph that is symmetrical about a vertical line that passes through a point on the curve called the vertex. The vertex will either be the highest point (maximum) or the lowest point (minimum) on the curve. The max-min condition is determined by the direction in which the parabola opens or, in other words, by the sign value of the coefficient, a. Thus, when the parabola opens upward, the vertex is the lowest point of the curve, and the same is true when the parabola opens downward, the vertex is the highest point on the curve. So, how do we find the vertex for any parabola?

 

First, the vertex is a point on the parabola, and, thus, is an ordered pair. We can find the vertex in one of two algebraic ways. If we know the function is in vertex form, as in #2 above, the vertex point can be read directly from the function:

 

y = 1.2(x + 3)2 + 7  tells us the location of the vertex is (– 3 , 7). Notice that the vertex form in #2 has a negative sign in the parentheses. This sign says to take the opposite sign of whatever is given in the parentheses when we write the vertex point.

 

Now suppose the function is in standard form:  y = 2x2 – 4x + 3. How do we find the vertex for this parabola? The vertex for a parabola whose rule is in standard form is as follows:

 

X =  – b          Y = f(– b )

                                                                                      2a                   2a

 

 

Calculating:      x = (-(-4))/(2(2)) or x = 1   So, y = 2(1)2 -4(1) +3 or 1.  The vertex is (1, 1). There are calculator methods that we will explore in class.

 

 

Intercepts: X and Y

 

     Let’s start with the y-intercept because it is the easiest. Recall that the y-intercept requires x = 0.

 

y = 2x2 – 4x + 3, in standard form, immediately yields the y-intercept of (0, 3). (Simply replace x with 0 in the function at the left.)

 

y = 1.2(x + 3)2 + 7, in vertex form, requires a little more work and care to get the vertex. Again, replace x with 0, and we have the following expression:

                                  1.2(3)2 + 7 = 1.2(9) + 7 =17.8  So, the vertex is (0, 17.8). In both cases, the vertices are minimums because the parabolas open upward.

 

 

     The x-intercepts are found, as we know, by setting y = 0 and solving for the x values. In the quadratic function, we use the quadratic formula for this process.  This formula is difficult to type, but here we go:

 

 

                                           X =  – b + sqrt(b2 – 4 a c)     and      X = – b – sqrt(b2 – 4 a c)    

                                                              2 a                                                 2 a

 

b2 – 4 a c is called the discriminant, and it gives us some vital information about the number of intercepts the quadratic function has. Since the discriminant is under the square root, we must consider the properties of where the square root is defined. We know that the value must be > 0. How does this affect the number of intercepts?

 

b2 – 4 a c < 0 tells us there are no intercepts.

 

b2 – 4 a c = 0 tells us there is one intercept (The vertex of the parabola sits on the x-axis).

 

b2 – 4 a c > 0 tells us there are two intercepts.

 

For y = 2x2 – 4x + 3, how many intercepts are there?  b2 – 4 a c = (-4)2 – 4 (2)(3)= -8  θ no intercepts.

So, there is no need to even begin calculations because the intercepts do not exist!

 

When the function does have intercepts, substitute the values for a, b, and c in the above formulas. You do not need to simply radical forms if the square is not perfect. However, if the square is perfect such as 36 or 81, these should be simplified and the arithmetic done to find the single values. Remember that x-intercepts are ordered pairs, and I expect you to report them this way.