179k views
0 votes
Which one of these is correct? Could you explain

Which one of these is correct? Could you explain-example-1
User DeKaNszn
by
7.5k points

1 Answer

2 votes
Because we have a x² leading term, this inequality is a quadratic and can be written in standard form.

Quadratic Standard Form:

ax²+bx+c=0

Replace = sign with <:

ax²+bx+c<0

So, we want to get 0 on the right side of the inequality to be in standard form.

To do this, we’ll add 18 to both sides:

x²-9x+18<0

This cancels out the -18 on the right, making it 0 because -18+18=0, and of course to keep the equation equal, we must add 18 to the left side as well.

Now, let’s understand the inequality. In standard form, a quadratic is equal to 0 to find the zeros. This is so that we can find what x-values will produce an output (y-value) of 0, as a y-value of 0 means no vertical change. It shows what x-values intercept the x-axis, which are the zeros (or roots) of the quadratic. Therefore, this inequality means: what x-values are less than 0, or go below the x-axis. To find this, we will factor the quadratic to see what x-values produce y-values less than 0.

Factor:

We know that -3-6=-9 and -3•-6=18

(x-3)(x-6)<0

So, the zeros of the function are x=3, 6. However, for x-values 3 and 6, the output is 0 since these values are x-intercepts. The inequality states less than 0, so we need look at x-values that produce a negative output.

So, if x=3, 6, then x>3 and x<6. Let’s look at the graph:
Which one of these is correct? Could you explain-example-1
User Vilsad P P
by
7.1k points