Suitable range of values for x

[Richie Smash]

Homework Statement

There is a function y= 3+x-2x²

X values are as follows: -3, -2, -1, 0, 1, 2, 3.
Y values are as follows:-18, -7, 0, 3, 2, -3,-12.

Each x value corresponds to the y value below it after going through the function.

I have plotted a graph using a scale of 1 cm on the Y axis and a scale of 2 cm on the X axis, I have obtained a Parabola opening downwards.

Now the Question says, '' Using your graph or otherwise, determine the range of values of x for which x -2x²>-6.

Homework Equations

The Attempt at a Solution

Now, I guess I can do this by substituing every x value into that inequality... but is there a faster way to do this

But in doing so I got -2 is less than or equal to x which is less than or equal to 2
Also how can I look at my graph and know this?

 

[scottdave]

Try plotting y = x - 2x² and compare to the original

 

[Richie Smash]

Ok, I have done that and obtained another downwards facing parabola, underneath the previous one.

 

[Richie Smash]

Here is a picture.

 

[Ray Vickson]

Homework Statement

There is a function y= 3+x-2x²

X values are as follows: -3, -2, -1, 0, 1, 2, 3.
Y values are as follows:-18, -7, 0, 3, 2,-3,-12.

Each x value corresponds to the y value below it after going through the function.

I have plotted a graph using a scale of 1 cm on the Y axis and a scale of 2 cm on the X axis, I have obtained a Parabola opening downwards.

Now the Question says, '' Using your graph or otherwise, determine the range of values of x for which x -2x²>-6.

Homework Equations

The Attempt at a Solution

Now, I guess I can do this by substituing every x value into that inequality... but is there a faster way to do this

But in doing so I got -2 is less than or equal to x which is less than or equal to 2
Also how can I look at my graph and know this?

The graph is a parabola opening downward, so for very large |x| (i.e, large positive or large negative) the graph

Homework Statement

There is a function y= 3+x-2x²

X values are as follows: -3, -2, -1, 0, 1, 2, 3.
Y values are as follows:-18, -7, 0, 3, 2,-3,-12.

Each x value corresponds to the y value below it after going through the function.

I have plotted a graph using a scale of 1 cm on the Y axis and a scale of 2 cm on the X axis, I have obtained a Parabola opening downwards.

Now the Question says, '' Using your graph or otherwise, determine the range of values of x for which x -2x²>-6.

Homework Equations

The Attempt at a Solution

Now, I guess I can do this by substituing every x value into that inequality... but is there a faster way to do this

But in doing so I got -2 is less than or equal to x which is less than or equal to 2
Also how can I look at my graph and know this?

Since the parabola opens downward, then for large positive x or large negative x the graph of $y =x - 2 x^2$ will lie below the line $y = -6$. Therefore, there must be some lowest value $x_1$ and largest value $x_2$ of $x$ that gives $x - 2 x^2 >-6$. These will occur where $x - 2 x^2 = -6$. This is a simple quadratic equation, which you can solve using the usual "quadratic root" formula. There are two roots: the smaller root will be $x_1$ and the larger root will be $x_2$.

 

[Nguyen Son]

Hi Richie Smash!
So this is your graph. The dots are the given values you wrote above.
"Determine the range of values of $x$ for which $x-2x^{2}>-6$"
You can see that $x-2x^{2}>-6$ means $y>-3$ (just plus 3 both side, nothing change)
Look at your graph, find every $x$ values which make $y>-3$ $\leftarrow$ just draw the line $y=-3$ then see which values $x$ "above" $y=-3$ line (I see 3 points)
Attention, do not obtain values $x$ where $y=-3$ because $x-2x^{2}>-6$, not $x-2x^{2}\geq-6$
That's the way you can use your graph, hope my guide can help you

 

[Richie Smash]

The graph is a parabola opening downward, so for very large |x| (i.e, large positive or large negative) the graphSince the parabola opens downward, then for large positive x or large negative x the graph of $y =x - 2 x^2$ will lie below the line $y = -6$. Therefore, there must be some lowest value $x_1$ and largest value $x_2$ of $x$ that gives $x - 2 x^2 >-6$. These will occur where $x - 2 x^2 = -6$. This is a simple quadratic equation, which you can solve using the usual "quadratic root" formula. There are two roots: the smaller root will be $x_1$ and the larger root will be $x_2$.

Hi Richie Smash!
So this is your graph. The dots are the given values you wrote above.
"Determine the range of values of $x$ for which $x-2x^{2}>-6$"
You can see that $x-2x^{2}>-6$ means $y>-3$ (just plus 3 both side, nothing change)
Look at your graph, find every $x$ values which make $y>-3$ $\leftarrow$ just draw the line $y=-3$ then see which values $x$ "above" $y=-3$ line (I see 3 points)
Attention, do not obtain values $x$ where $y=-3$ because $x-2x^{2}>-6$, not $x-2x^{2}\geq-6$
That's the way you can use your graph, hope my guide can help you

Hi Ray vickson, I solved the quadratic and obtained
x= -1.5
x= 2

Hi Nyugen Son, I'm not quite sure about your method?

 

[Richie Smash]

I understand the solving the Quadratic equation, but How do I check my answers on the graph to determine the range?

 

[scottdave]

Try graphing y = -6 on top of your parabola graph

 

[Richie Smash]

When I then drew the line y = -6, I found the range which matches up with the quadratic equation I solved for the parabola x-2x².

However the range with the line y= -6 for the parabola y= 3+x-2x² was not the correct range.

I was only when I drew the line y= -3 did the range for the parabola y= 3+x-2x that the range was identical to the one found using the line y=-6 for the parabola x-2x².

I think I understand now as I can see the ranges clearly on my graph, what I don't understand is why does changing y = -6 to y = -3 make the original parabola also satisfy the inequality?

 

[Ray Vickson]

I understand the solving the Quadratic equation, but How do I check my answers on the graph to determine the range?

You don't need to (but doing so cannot hurt).

The smallest root $x_1 = -1.5$ is a dividing point between the regions where $x -2x^2 < -6$ and $x - 2 x^2 > -6$. Since for large negative $x$ we have $x - 2x^2 < -6$ (parabola below the line at -6), all points to the left of $x_1$ give $x - 2x^2 < -6$ and points a bit to the right of $x_1$ have $x - 2x^2 > - 6$. That continues to happen until $x$ grows large enough to hit $x_2 = 2$, which is the second dividing point between $x - 2x^2 > -6$ and $x - 2 x^2 < -6$. Therefore, all points between $x_1$ and $x_2$ satisfy the required inequality.