Question

Find the x-intercepts for the parabola defined by the equation below.

y= 2^2 + 2x - 4

a. -4,0 and 2,0
b. -2,0 and 1,0
c. 0,-2 and 0,1
d. 0,-4 and 0,2

Answer 1

b.

Explanation:

x-intercepts are found by factoring. We will use standard factoring here since this one is straightforeward and has real zeros as its solutions.

In our equation,

a = 2

b = 2

c = -4

The rules are to take a * c and then find the factors that number, determine which combination of those factors will give you the linear term (the term with the single x on it), and rearrange those signs accordingly. Let's start with that:

Our a * c is 2 * -4 = -8.

We need the factors of |-8|: 1,8 and 2,4

Some combination of those factors needs to give us a +2x. 2,4 will work as long as the 4 is positive and the 2 is negative.

Now we put them back into the equation, the absolute value of the larger number first:

`2x^2+4x-2x-4=0`

Now group the terms in sets of 2 without moving any of them around:

`(2x^2+4x)-(2x-4)=0`

In each set of parenthesis, pull out what is common to both terms. In the first set, the 2x is common, and in the second set, the 2 is common:

`2x(x+2)-2(x+2)`

Now what is common between both terms is the (x + 2), so pull that out, grouping what is remaining in its own set of parenthesis:

`(x+2)(2x-2)=0`

To find the zeros, remember that the Zero Product Property tells us that for that equation above to equal zero, one of those factors has to equal zero, so:

x + 2 = 0 or 2x - 2 = 0. Solve both for x:

x = -2 so the coordinate is (-2, 0)

2x - 2 = 0 and

2x = 2 so

x = 1 so the coordinate is (1, 0)