Algebra Know-It-ALL

512 Review Questions and Answers

In this quadratic, a= 1 and b=−2. Therefore

xv=−(−2) / (2 × 1)

= 2/2

= 1

Plugging this in and working out the arithmetic using the product of binomials, we get

yv= (xv+ 2)(xv− 4)

= (1 + 2)(1 − 4)

= 3 × (−3)

=− 9

Therefore, (xv,yv)= (1, −9). This vertex is an absolute minimum because, as we found in

Answer 24-5, the parabola opens upward.

Question 24-7

Based on the information in Answers 24-5 and 24-6, how can we sketch an approximate

graph of the quadratic function stated in Question 24-5?

Answer 24-7

We have found that the zeros of the function are x=−2 and x= 4. The points representing

them are at (−2, 0) and (4, 0). The vertex point is at (1, −9), and the graph is a parabola that

opens upward. Knowing all this, we can sketch the graph as shown in Fig. 30-1.

Question 24-8

Suppose we come across the following quadratic function in binomial factor form, where x is

the independent variable and y is the dependent variable:

y=− 3 x^2 + 7 x− 11

Does the parabola representing this function in Cartesian or rectangular coordinates open

upward or downward? How many real zeros does the function have?

Answer 24-8

The parabola opens downward because the coefficient of x^2 is negative. To determine how

many real zeros the function has, we can evaluate the discriminant d. Once again, recall the

general polynomial standard form for a quadratic function:

y=ax^2 +bx+c

Here, we have a=−3,b= 7, and c=−11. Therefore

d=b^2 − 4 ac

= 72 − 4 × (−3)× (−11)

= 49 − 132

=− 83