This can be rearranged to get
ax^2 +bx+ (c−k)= 0which we can solve with the quadratic formula. The roots of this quadratic, which we’ll call p and q, are
thex-values of the two points where the parabola intersects the horizontal line y=k. By doing all this, we
learn the coordinates of three points that lie on the parabola. Those points are (p, k), (q, k), and (r, 0 ). If
we’ve chosen the value of k wisely (made a lucky guess), we’ll have three well-separated points, and we’ll
have an easy time drawing an approximation of the parabola.
Here’s a challenge!
Look at the following quadratic function:
y=− 4 x^2 + 12 x− 9Find the zero or zeros. Determine whether the parabola for this function opens upward or downward.
Find the x-value and the y-value of the extremum. Then find two more points so the curve can be drawn.
Finally, draw an approximation of the curve.
Solution
We can find the zero or zeros of the function by applying the quadratic formula to the equation
− 4 x^2 + 12 x− 9 = 0In the general polynomial equation
ax^2 +bx+c= 0we have a=−4,b= 12, and c=−9. When we plug these into the quadratic formula, we get
x= [−b± (b^2 − 4 ac)1/2] / (2a)
= {− 12 ± [12^2 − 4 × (−4)× (−9)]1/2} / [2 × (−4)]
= [− 12 ± (144 − 144)1/2] / (−8)
= (− 12 ± 02 ) / (−8)
=−12/(−8)
= 3/2
The equation has only one root, so the function has only one zero. If we call it r, then we have
r= 3/2, and we know that the point (3/2, 0) is on the parabola. We also know that this point repre-
sents the extremum. The value of a, the coefficient of x^2 in the polynomial, is negative. That means
the parabola opens downward, so (3/2, 0) is the absolute maximum, and the parabola “hangs from
thex axis.” We can choose a negative number k and set it equal to y, getting a horizontal line that
crosses the parabola twice. Let’s try −5 for k, as shown in Fig. 24-7. The equation of the line is y=−5.
One Real Zero 405