Solve Absolute Parabola: Sketch |x^2 - 2x - 3| Graph

[TsAmE]

Homework Statement

Sketch the graph of |x^2 - 2x - 3|

Homework Equations

None

The Attempt at a Solution

|x^2 - 2x - 3| =

{ x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
{-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1

if x < -1 then y = -x^2 + 2x + 3

if -1 <= x < 3 then y = (x^2 - 2x - 3) + (-x^2 + 2x + 3)
= 0

if x >= 3 then y = x^2 - 2x - 3

I wrote this in a test and I got the part in bold wrong. I don't know what I did wrong as I have used this method of showing the positive and negative values of an absolute graph before and it has worked.

 

[Mark44]

Homework Statement

Sketch the graph of |x^2 - 2x - 3|

Homework Equations

None

The Attempt at a Solution

|x^2 - 2x - 3| =

{ x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
{-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1

The inequalities at the end of each line are wrong.
x >= 3 OR x >= -1 is the same as saying x >= -1. By "the same as" I mean "equivalent to."
x < 3 OR x < -1 is the same as saying x < 3.

if x < -1 then y = -x^2 + 2x + 3

No, this isn't true.

if -1 <= x < 3 then y = (x^2 - 2x - 3) + (-x^2 + 2x + 3)
= 0

This isn't true, either.

if x >= 3 then y = x^2 - 2x - 3

I wrote this in a test and I got the part in bold wrong. I don't know what I did wrong as I have used this method of showing the positive and negative values of an absolute graph before and it has worked.

Look at the graph of y = x^2 - 2x - 3. This graph crosses the x-axis at (-1, 0) and (3, 0). These points determine three intervals: (-inf, -1), (-1, 3), and (3, inf). On one of these intervals the graph if y = x^2 - 2x - 3 dips below the x-axis.

For y = |x^2 - 2x - 3|, the part that was below the x-axis is reflected across. The other two parts don't change.

 

[TsAmE]

if x < -1 then y = -x^2 + 2x + 3

if -1 <= x < 3 then y = (x^2 - 2x - 3) + (-x^2 + 2x + 3)
= 0

if x >= 3 then y = x^2 - 2x - 3

The funny thing is that I plotted those equations in bold (all above positive y-axis cause of absolute) and got the graph right, but I don't understand why my reasoning would be wrong as I stated the relevant intervals: x < -1, -1 <= x < 3 and x >=3.

|x^2 - 2x - 3| =

{ x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
{-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1

I got the first bold part by saying (x - 3)(x + 1) >= 0
and the second bold part by saying (x - 3)(x + 1) < 0
and just solving for both x values. What would be the correct inequality(s) to right?

 

[Mark44]

The funny thing is that I plotted those equations in bold (all above positive y-axis cause of absolute) and got the graph right, but I don't understand why my reasoning would be wrong as I stated the relevant intervals: x < -1, -1 <= x < 3 and x >=3.

Those are the right intervals.

|x^2 - 2x - 3| =

{ x^2 - 2x + 3 x^2 - 2x - 3 >= 0 x >= 3 OR x >= -1
{-x^2 + 2x + 3 x^2 - 2x - 3 < 0 x < 3 OR x < -1

I didn't notice earlier, but the first inequality in each line has extra terms in it. Probably a copy and paste error.

I got the first bold part by saying (x - 3)(x + 1) >= 0

The solution to this inequality is x < = -1 OR x >= 3. This represents two separate intervals.
You have x >= 3 OR x >= -1, which as I said before is the same as saying x >= -1, which is only one interval.

and the second bold part by saying (x - 3)(x + 1) < 0
and just solving for both x values. What would be the correct inequality(s) to right?

The solution to this inequality is -1 < x < 3. You had x < 3 OR x < -1, which as I said before, is the same as x < 3.

Your problem seems to be solving quadratic inequalities. You should go back and review the section in your book that discusses this type of inequality.

 

[TsAmE]

Oh ok I understand where I went wrong, I didnt apply the use a number line with critical values to get the interval. Thanks for the help :)