Algebra Know-It-ALL

402 Graphs of Quadratic Functions

to get a clear picture of the parabola based on their locations. But we can find the y-intercept to help us

draw the curve. When we plug the value 0 in for x, we get

y=x^2 − 3 x+ 2

= 02 − 3 × 0 + 2

= 0 − 0 + 2

= 2

This tells us that (0, 2) is on the curve. It is also shown in Fig. 24-4.

Here’s a trick!

The graph of a quadratic function is bilaterally symmetric with respect to a vertical line passing through the

vertex. That means the right-hand side of the curve is an exact “mirror image” of the left-hand side. In the

challenge we just finished, this fact can be useful. Once we’ve drawn an approximation of the left-hand

side of the curve using the points (3/2, −1/4), (1, 0), and (0, 2), we can fill out the right-hand side without

having to find another distant point there.

One Real Zero

When a quadratic equation has one real root, its quadratic function has one real zero. When we

graph such a function, it has a single point in common with the independent-variable axis.

Parabola opens upward

Figure 24-5 shows a generic graph of a quadratic function of x with one real zero r. At the

point (r, 0), the parabola is tangent to the x axis. Here, “tangent” means that the curve just

brushes against the axis, touching it at a single point. If the quadratic function is

y=ax^2 +bx+c

thena > 0 because the parabola opens upward. The root of the equation

ax^2 +bx+c= 0

can be found using the quadratic formula. Because there is only one root, the discriminant in

the quadratic formula must be 0. That means

b^2 − 4 ac= 0

Therefore

r= [−b± (b^2 − 4 ac)1/2] / (2a)

=−b/(2a)