Algebra Know-It-ALL

We could shrink it by making each division represent, say, 5 units on both axes. But then the
solutions wouldn’t show up well; they’d be too close together.
In situations like this, the best approach is to use rectangular coordinates, but not strict Car-
tesian coordinates. Let’s make each increment on the x axis represent 1 unit, and each increment
on the y axis represent 5 units. When we make the axis increments different in size, we distort
the slopes and contours of the lines and curves, but that’s not important here. Our goal is only
to get a good fit for the ranges of the values we determined when we created Table 28-1, and a
clear picture of how the graphs intersect. To begin, we plot the two solution points.

Finally, we plot the rest

Once we’ve plotted the solution points as dots on the coordinate grid, we can locate and plot
the rest of the points in the table. If we do this on paper, we can use pencil and draw the points
lightly, so we can erase them later.
As we draw the points, we must remember the increments we’ve chosen for each axis.
They aren’t the same, and it’s easy to get them confused. It’s also important to keep track of
which graph goes with which function!
We can draw the points for one graph with a pencil, fill in its line or curve with a pen,
draw the points for the second graph with a pencil, fill in its line or curve with a pen, and then
erase all the penciled-in points when the ink is dry. If we’ve chosen the axis increments wisely,
the solutions will be near the center of the coordinate area, and the other points will be scat-
tered fairly well over the rest of it. Fig. 28-1 shows the final result.

Here’s a challenge!

By examining Fig. 28-1, describe how the linear function

y= 2 x+ 1

(shown by the solid line) can be modified to produce a system with no real solutions, assuming that the qua-
dratic function (shown by the dashed curve) stays the same, and assuming the slope of the line stays the same.

Solution

If we change the linear function so its graph doesn’t intersect the parabola, then the resulting two-by-two
system will have no real solutions. Imagine moving the straight line downward in Fig. 28-1. As we do this,
the solution points get closer together, eventually merging. At the moment the two points become one,
the system has a single real solution with multiplicity 2.
The linear function is expressed in slope-intercept form, so we can see that the line has a slope of 2 and
ay-intercept of 1. Each division on the y axis is 5 units, so we can see that the quadratic function has an
absolute minimum of approximately −5. Suppose we change the linear function so the y-intercept is −10.
That will put the line completely below the parabola, and will give us the two-by-two system

y= 2 x− 10

and

y=x^2 +x− 5

Linear and Quadratic 465