Algebra Know-It-ALL

514 Review Questions and Answers

In this function, a=−3 and b= 7. Therefore

xv=−7 / [2 × (−3)]

=−7/(−6)

= 7/6

Plugging this in and working out the arithmetic using the function, we get

yv=− 3 xv^2 + 7 xv− 11

=− 3 × (7/6)^2 + 7 × (7/6) − 11

=−49/12+ 49/6 − 11

=−49/12+ 98/12 − 132/12

= (− 49 + 98 − 132) / 12

=−83/12

Therefore, (xv,yv)= (7/6,−83/12). This vertex is an absolute maximum, because the parabola

opens downward.

Question 24-10

Locate two points on the graph of the function stated in Question 24-8, other than the vertex.

How can we sketch an approximate graph of the function?

Answer 24-10

We can locate two points by plugging in a value of x somewhat smaller than xv, and another

value of x somewhat larger than xv. Let’s try x 1 = 0 and x 2 = 2. In the first case, we have

y 1 =− 3 x 12 + 7 x 1 − 11

=− 3 × 02 + 7 × 0 − 11

= 0 + 0 − 11

=− 11

The first non-vertex point is (x 1 ,y 1 )= (0, −11). In the second case,

y 2 =− 3 x 22 + 7 x 2 − 11

=− 3 × 22 + 7 × 2 − 11

=− 12 + 14 − 11

= 2 − 11

=− 9

The second non-vertex point is (x 2 ,y 2 )= (2, −9). Now we know these things:

• The vertex point is (7/6, −83/12)


• The parabola contains two other points (0, −11) and (2, −9)


• The parabola opens downward