Minimum or maximum value of a parabola - Texas Instruments TI-84 Plus Silver Edition Graphic Calculator

Domain and range are concepts that many students have trouble with, so you are not alone!

Domain is the x values of the relation.
Range is the y values of the relation.

If you have trouble remembering which one is which , like I do, D comes before R, just like x comes before y;)

To determine if a relation is a function, there are several tests that you can use. For every x value, there can be one and only one y value. If there is more than one y value for any x value, it is not a function.

Another test is the vertical line test. If a vertical line only goes through one point on the line, it is a function. If it goes through two points on the line, it is not a function.

As an example, let's look at y = (x-3)^2 + 6.

This parabola is in the form, y=a(x-h)^2 + k, where a indicates a stretch or compression, and whether the parabola opens up or down. If a is positive, it opens up, and the y value of the vertex represents a minimum. If a is negative, it opens down, and the y value of the vertex is a maximum. The values h and k are the x and y values of the vertex.

In this case, a is one, so the parabola opens up, with a minimum value of 6.''

Now back to the domain and range.

Since we can use any value for x in the equation, for the domain, x is an element of real numbers, sometimes written (xER).

For the range, y can only be greater than or equal to the minimum. So the range is y is an element of all real numbers, such that y is greater or equal to 6, sometimes written (yER'y>=6). Sorry I couldn't find the greater than or equal to character;)

Good luck.

Paul

Set the windows dimensions to Standard (Xmin=-10, Xmax=10, Ymin=-10, Ymax=1: Both are parabolas, and the interesting part is the neighborhood of the extremum (maximum or minimum) and the location of the zero.
The third curve varies very fast, so you will have to reduce the interval Xmin, Xmax. to see the neighborhood of the origin, where the minimum is. (Minimum is at (0,0).

The maximum (or minimum) occurs when the first derivative is zero. In this case the maximum y value is 16 when x is 3.

If this is homework, be sure to show your work.

The maximum value of the sine function is positive one, the minimum value is negative one. Within any limited range the maximum and minimum will fall between the two extremes.
If you have a graphing calculator, the easiest way to to find the maximum and minimum values of a sine curve is to plot it. Some calculators will even automatically find the max and min values. Others you'll have to eyeball it.
If you don't have a graphing calculator, you can calculate the value at a series of points and hone in the max and min values.
If you need further assistance, please reply to this post and specify the make and model of your calculator.

Follow these steps: graph the parabola and after that 2nd then CALC keypad. To find extreme value of the function select 3: min or 4: max options. Select the function and set left bound, right bound(using left and right arrows) by pressing ENTER. You can see images bellow for all these steps for the y=-2x^2+5x+3


1. step: graphing parabola



2. step: left or lower bound



3. step: right or upper bound



4.step: extreme value of the function

Hello Snsimpson



Here are all the specs for the Windstar.


FRONT ALIGNMENT (CURB HEIGHT WITH 1/2 TANK OF FUEL)
Camber

Nominal -0.25°
Minimum -0.75°
Maximum +0.25°
Split 0.0°±0.5° Caster
Right

Nominal +3.80°
Minimum +3.05°
Maximum +4.55°
Split -0.5°±0.5° Left

Nominal +3.30°
Minimum +2.55°
Maximum +4.05°
Split -0.5°±0.5° Total Toe [1]

Nominal -0.10°
Minimum -0.40°
Maximum +0.20° Clear Vision [2]

Nominal -0.2°
Minimum -2.8°
Maximum 3.2° REAR ALIGNMENT (CURB HEIGHT WITH 1/2 TANK OF FUEL)
Camber

Nominal 0.0°
Minimum -0.3°
Maximum +0.3° Total Toe [1]

Nominal -0.06°
Minimum -0.40°
Maximum +0.28° [1] Positive value specifications for toe-in, negative value specification for toe is toe-out. [2] Negative value specification for clear vision is counterclockwise.

The actual graph depends on the constants (parameters) that are in the equation. Try playing with the dimensions of the window by zooming in or zooming out.
The most interesting part of the graph is the region near the maximum or minimum.
Here is a graph of the function y=X^2-5X 13 when drawn with standard window dimensions

Not very revealing. However if you press the Zoom key and select [2:ZoomIn] you see this (no changes) but the calculator is ready to accept your input concerning the new center of the graph.

At this stage, use the Arrow keys to move the center of the graph to another point as in the next screen capture.


Notice the new center as a sign on the screen and its coordinates at the bottom. By pressing [ENTER] you get this

If you keep Zooming in you may end up having something that does not look like a parabola anymore.


To summarize: Use the Zoom In or Zoom Out functions or find where the minimum/maximum is, center the graph on that point and modify the windows dimensions (Xmin, Xmax, Ymin, and Ymax) so as to get as much of the interesting part of the graph as possible.

If the parabola has its concavity turned downward and it maximum value is lower than 0 then the value of the functions are always negative (never reach 0).
Similarly, if the parabola has it concavity turned upward, and its minimum value is positive, then all the values of the functions are positive (never reach zero). Thus if y is never equal to zero the function has no x intercepts.
The concavity is called by some people the mouth.

Parabolas
Open Y= editor and type in the two functions

The calculaus functions are accessible by pressing [2nd][TRACE] to open the CALCulate menu options. For the gradient (I think you mean the derivative) use option 6:dy/dx. But first choose the point where you want it calculated (use cursor to move along the curve) and press ENTER. The value of the deivative will be calculated at the chosen point.


The vertex of the parabolas are maxima. Thus you must use option 4:maximum

You will be prompted for a left bound. Move cursor to the left of the maximum (not too far) and press [ENTER]. A fat arrow is displayed on the graph that shows the left limit of the interval. You will be prompted for a right bound. Move cursor along the curve to the right of the the vertex. Press ENTER. A seconf fat arrow will be displayed to show the right limit of the interval.

You will be prompted for a guess of the maximum. Move cursor newar the max or enter a numerical value and press ENTER.



The location of the vertex is displayed (X and Y values)

I have no idea what you mean by the equation of symmetry

Intercept.

Hello,
When you are use the calculator to find a maximum or a minimum, you are supposed to give the calculator a range of values between which to search for i (interval where the max or min lies)
1.Look at the graph to locate approximately the position (the x coordinate) of the minimum.
2. Supply the left limit of x ( to the left of the minimum) Press [ENTER]
3. Supply the right limit of x (to the right of the minimum). Press [ENTER]
3. Supply an estimate of the X-value of the minimum.


After a moment the calculator displays the X and Y values of the maximum. The better your estimate, the faster the maximum will be found.


Hope this will help you set your problem correctly.