by:

Brandt Hacker

For this assignment we are asked to determine what the equation 2x + b = 0 has in relation to quadratic functions.  For the remaining graphs and equations we will replace b with y in order to show the graphs on the xy-plane, which is recognized by the graphing program we are using.

We will first look at the graph of a quadratic equation with both a and c fixed at the value of 1, while we allow the coefficient of the middle term to vary.  Here is the first example of our replacing b with the variable y in order to show how the change in b affects the graph.

By fixing the values of a and c in the quadratic equation, the graph above gives us all of the roots of the equation .  Below we will overlay various values of y to highlight this concept.  First we look at an equation with two real roots:

Above we see that when b = 4, the quadratic equation has two real roots, both of which are located at the intersection points of the equation  and y = 4.

Next we look at an equation with no real roots:

In the situation we see above, there is no intersection of the graphs, and therefore the quadratic equation when b = 1 has no real roots.  This is seen from the fact that the graph does not ever hit the x-axis.

Lastly we will look at a situation with exactly one real root:

In the graph shown above, we have a situation where the line y = 2 is tangent to the graph of the roots of the quadratic.  Because our horizontal line at y = 2 touches the graph of  at exactly one point, we know that the graph of the quadratic  has exactly one real root, the point at which the graph hits the x-axis.  Algebraically we can show this by taking the quadratic equation and factoring it,  is an equation that lends itself nicely to demonstrating this.

Now going back to the original question, how does 2x + b = 0 relate to what we have discussed thus far?  If we lay the graph of 2x + y on top of our previous graph, we begin to see how the two are related.

Here we see that the equation 2x +y = 0 gives us two intersection points with the equation , one at the point (-1, 2) and another at the point (1, -2).  When we take the y-values 2 and -2 and plug them back into the equations for a quadratic as values of b, we get both the red and purple graphs, both of which have one real root and are tangent to the x-axis.  It appears then that the equation 2x + b = 0 when graphed in the xb-plane gives us the values of b that will create equations with exactly one real root given a fixed value for a and c.

If we change the fixed value of c from 1 to -1, the graph changes considerably.

Here we have two equations with no intersection. Based on what we saw before, this makes sense because of the fact that the new y-intercept for any quadratic of the form  will be -1.  As we saw in assignment 2, the positive  term will cause the parabola to open up.  With a negative y-intercept it is not possible to have a point of tangency with the x-axis with a parabola that opens up.

WeÕve now seen how the equation is useful in determining the value of b that will produce an equation with one real root, if such an equation exists.  LetÕs illustrate this concept one last time with an equation that does not factor quite as nicely as the one that we looked at previously.

In the graph above, the intersection point is not as clearly seen as one with the form .  We set up the following equation to find the intersection point

After solving for y in both equations we set them equal to one another in an attempt to solve the system.

We can then take this value of x and plug it back in to either of the original equations to find y.  When this is done we see that

When this is plugged back into a quadratic for the value of b, the resulting graphs are seen in purple and red below.