Constants Examples

Example 1:


Several values of x and their corresponding values of y are shown in the table. A linear equation that represents the relationship shown in the table is x = k, where k is a constant. What is the value of k ?

This problem is as straightforward as it seems – constants don’t change, so they are saying x is equal to some number. What number? It’s in the table – no matter what y is, x is always going to be 2. If we were to graph this, it would just be a vertical line.

Example 2:

The linear function f is defined by f (x ) = cx + d, where c and d are constants. If f (50) = 27,000 and f (100) = 38,000, what is the value of c ?

Here we have two constants, which, remember, are just two numbers we don’t know yet but will shortly. If we look closely we can see that


looks an awful lot like


And in fact, we do have a linear equation here. They are asking us for the slope of this line, but calling it c instead of m. A common SAT trick!

In our functions unit we worked through problems like this, where we plugged in a number and got back another number.

Here, they tell us if we plug in 50, we get back 27,000, and if we plug in 100, we get back 38,000. Remember we plug in for x and get back y. So, we have two ordered pairs! (50, 27000) and (100, 38000). Also recall from our unit on slope that we can calculate the slope from two points, using (frac{y_1-y_2}{x_1-x_{2_{}}}=m). Here we just plug in our two points and get back the slope, which is also equal to the constant c. If they asked us for d, the y-intercept, we could solve by plugging in one of our points back in along with our newly found slope.

Example 3:


For the quadratic equation shown, b is a constant. If the equation has no real solutions which of the following must be true?

A) (b^2< 60)
B) (b^2>60)
C) (b<0)
D) (b>0)

b is just some number we don’t know yet!

We remember from our unit on the quadratic formula and the discriminant that for a give quadratic, if the discriminant is positive it will have 2 solutions, if it is 0 there will be 1 solution, and if it is negative there will be no solutions. Here, the easiest way to solve this problem will be to use that rule.

First, we define our a, b, and c terms.

a = 3

b = b

c = 5

Plugging this into our discriminant formula give us




(b^2) must be less than 60 here in order for our discriminant to be negative.