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For this assignment we will be examining the graphs of:

for multiple values of a,b, and c.

 

For our first exploration let's allow a to vary from -5 to 5 and keep b and c fixed at 1.

Observations?

• When a=0 we simply have the equation y=bx+c which for b,c=1 gives us the equation y=x+1. This is a line with x-intercept at (-1,0) and y-intercept at (0,1)
• For negative values of a the graph is a parabola pointing down and the positive values for a invert this parabola across the x-axis
• No matter what value we had for a, the graph always intersects the y axis at (0,1)
• as the absolute value of a increases the parabola gets "narrower"

Next let b vary from -5 to 5 and let a and c be fixed at 1

Observations?

• When b=0 we simply have the equation y=a(x^2)+c which for a,c=1 is just y=(x^2)+1. This is a parabola pointed upwards with a vertex at (0,1)
• When the values for b are negative the vertex of the parabola is to the right of the y-axis, when the values of b are positive this parabola is inverted across the y-axis which means the vertex is now on the left of the y-axis
• no matter what value we used for b the parabolas have a y-intercept at (0,1)
• notice that if the absolute value of b is 3 or greater then the parabola crosses the x-axis at two points, however for |b|=2 the parabola touches the x-axis only at one point and for |b|=1 or 0, the parabola does not cross the x-axis at all. What does this say about root values of the equation (x^2)+bx+1=0 ?

And finally let c vary from -5 to 5 but a and b remain fixed at 1

Observations?

• changing the value of c changes the y value for the parabola's vertex. As c decreases so does the y value of the vertex. However, note that c does not impact the x value of the parabola's vertex.
• if c is less than or equal to 0 then the parabola crosses the x-axis twice but if c>0 then the parabola does not cross the x-axis

 

Graphing in the xb plane

Let's go back to the discussion of the roots of:

The graph for (x^2)+bx+1=y for values of b ranging from -5 to 5 looked like this:

 

But what if we spun this whole thing on its head and graphed the equation in the xb plane instead of the xy plane?

Graphing the equation in the xb plane give us:

If we then graph any particular value of b, say b=5, we get a line parallel to the x-axis. Where this line intersects the graph of (x^2)+bx+1=0 we find the root values for that particular value of b.

See below for an example.

 

Let's look at the graphs in the xb plane with b equaling -5 to 5

From this graph it becomes clear that when b > 2 the equation (x^2)+bx+1=0 has two real positive roots and when b < -2 the equation has two negative real roots. For b=2 there is one positive real root and when b= -2 there is one negative real root. There are no real roots when -2 < b < 2.

 

What if we changed the value of c?

Recall the graph of (x^2)+bx+c=0 with b=1 and c varying from -5 to 5 in the xy plane:

From the above graph what do you imagine the roots are for the equation (x^2)+x-3 = 0 are?

Let's graph (x^2)+bx-3=0 and the line b=1 in the xb plane:

As you can see there are two real roots.

 

But if we change the equation to (x^2)+bx+3=0?

This shows us that there are no real roots for this equation.

 

Try on your own other values for c and b in the xb plane.

 

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