Algebra Know-It-ALL

682 Worked-Out Solutions to Exercises: Chapters 21 to 29

3. Because the parabola opens upward, we know that its vertex is an absolute minimum.
   To find the x-coordinate of this point, we average the two zeros:

xmin= (3 + 1/4)/2

= (12/4 + 1/4)/2

= (13/4)/2

= 13/8

We can calculate the y-coordinate of this point by plugging in 13/8 for x in the polyno-

mial form of the function:

ymin= 4 xmin^2 − 13 xmin+ 3

= 4 × (13/8)^2 − 13 × 13/8 + 3

= 4 × 169/64 − 169/8 + 3

= 676/64 − 1,352/64 + 192/64

= (676 − 1,352 + 192) / 64

=−484/64

=−121/16

The coordinates of the absolute minimum are (13/8, −121/16).

4. We know the two points where the curve crosses the x axis (representing the zeros of
   the function). The left-hand x-intercept point is (1/4, 0). The right-hand x-intercept
   point is (3, 0). We also know that the absolute minimum point is (13/8, −121/16). The
   graph, based on these three known points, is shown in Fig. C-2. On both axes, each
   increment represents 1 unit.


5. Remember the polynomial standard form for a quadratic function of a variable x, where
   we have coefficients a and b, and a stand-alone constant c. If we let the dependent
   variable be called y, then

y=ax^2 +bx+c

The function we’re interested in is

y= 7 x^2 + 5 x+ 2

In the polynomial, we have a > 0, so we know that the graph of the function is a parabola

that opens upward. When we examine the discriminant d, we find that

d=b^2 − 4 ac

= 52 − 4 × 7 × 2

= 25 − 56

=− 31

The fact that d is negative tells us that the quadratic equation

7 x^2 + 5 x+ 2 = 0