Graph to determing when greater than a number

[JSmith]

Create a graph in notebook software to determine when 6x³ - 5x² - 2x + 12 > 11. Show all of your work.

How do I graph this and show my work, I dont know how to graph this without using a graphing calculator. I could use a table to determine positive and negative intervals, however that doesn't tell me if the function is greater than 12, just if its greater than 0. How would you solve this????

 

[JSmith]

The instruction is clear; graph it using software.

Are you telling us that your teacher will not accept your work, if you use a graphing calculator to answer the exercise versus using some notebook software (whatever that is)? :?

Yes, drawing with notebook software is no different than hand drawing. So essentially I have to include a drawn graph in the answer.

 

[mmm4444bot]

Did you subtract 11 from both sides of the inequality, already?

Graph the polynomial using your calculator in a window with x from -2 to 2 and y from -3 to 3. Zoom in on the intercepts. Get some other obvious points on the graph.

Then, carefully copy onto graph paper.

 

[soroban]

Hello, JSmith!

This can be done without a calcuator.

\(\displaystyle \text{Graph: }\(x) \:=\:6x^3 - 5x^2 - 2x + 12 \: >\: 11\)

We have: .\(\displaystyle 6x^3 - 5x^2 - 2x + 1 \;>\; 0\)

We find that \(\displaystyle x=1\) is a zero of the polynomial.

Hence: .\(\displaystyle P(x) \:=\: (x-1)(6x^2 + x - 1) \:=\: (x-1)(2x+1)(3x-1)\)

The zeros (x-intercepts) are: \(\displaystyle 1,\:\text{-}\frac{1}{2},\:\frac{1}{3}\)

A cubic graph is shaped like this .\(\displaystyle _{\cup}\!^{\cap}\:\text{ or }\:^{\cap}\!_{\cup}\)

Plot a few points to determine its shape.

 

[srmichael]

Hello, JSmith!

This can be done without a calcuator.


We have: .\(\displaystyle 6x^3 - 5x^2 - 2x + 1 \;>\; 0\)

We find that \(\displaystyle x=1\) is a zero of the polynomial.

Hence: .\(\displaystyle P(x) \:=\: (x-1)(6x^2 + x - 1) \:=\: (x-1)(2x+1)(3x-1)\)

The zeros (x-intercepts) are: \(\displaystyle 1,\:\text{-}\frac{1}{2},\:\frac{1}{3}\)

A cubic graph is shaped like this .\(\displaystyle _{\cup}\!^{\cap}\:\text{ or }\:^{\cap}\!_{\cup}\)

Plot a few points to determine its shape.

Extending on Soroban's work, you can also plot the zeros on a number line (essentially the x-axis) and choose test points: one less than -1/2, one between -1/2 and 1/3, one between 1/3 and 1 and one greater than 1. Plug these test points into the factored form and determine which test point produces a value > 0.

 

[JSmith]

Thanks, I actually solved it last night using pretty much that exact method!

 

[mmm4444bot]

This can be done without a calcuator.

Of course! But then you are not following the given instruction to use software.