Each of these are very beneficial and can even save you work when it comes to minor calculations that will lead to a "big picture" like many of those used in junior and senior classes or standardized test.1: Value
Finding the value of y at a given x value on a graph can be important for many reasons. Letting the calculator solve for these values can also save you time and effort.
With the graph on the screen enter the calculate
menu and choose value
. This will then prompt you for an x-value. Enter the value using the numeric keys and press enter to see the corresponding y-value and it's position on the graph. The example is for y=2x and finding the value when x=4.2:Zero
Finding the zeros or the roots of a function is used in many different applications. Finding these can be tedious if you have to set an entire function equal to zero and then solve for x algebraically. To find these fast use the zero
function on the calculator.
Access the zero
function from the calculate
menu. The calculator will prompt you to move the cursor to the "Left-Bound" or an x-value to the left of where the root is. After pressing ENTER it will then ask you to do the same for the "Right-Bound" and press ENTER again. It will then as you for a "guess" (which isn't needed 99% of the time) so press ENTER again and the zero of the function will be displayed.
For this example I used the parabola shifted down by 2.3:Minimum
Sometimes it is helpful to find where the minimul value is on a curve. I started by graphing a simple parabolic curve.
Start by accessing your calculate menu and then choosing Minimum. The calculator will then ask you to find a left-bound point. Use the arrows to move your cursor to the left of where the minimum is and then press ENTER. Repeat this procedure for the right-bound point. The guess in most instances is trivial so press ENTER again to see the minimum value. 4:Maximum
By the same token it can be helpful to find where the maximum value is on a curve. I started by graphing a simple negative parabolic curve.
by accessing your calculate menu and then choosing Maximum. The
calculator will then ask you to find a left-bound point. Use the arrows
to move your cursor to the left of where the maximum is and then press
ENTER. Repeat this procedure for the right-bound point. The guess in
most instances is trivial so press ENTER again to see the maximum
When solving a system or equations or finding a common value for any reason it can be helpful to graph two equations and find where they intersect or meet.
Start by graphing both functions and then accessing the intersect feature under the calculate menu. It will then prompt you to place the cursor on the first curve. This is already done if there are only two graphs shown or use the arrow keys to get there. Then press ENTER and move the cursor to the second graph before pressing ENTER again.
It will again ask you for a guess but in most cases is is irrelevant so press ENTER again to see the point of intersection.6:Derivative (dy/dx)
Finding the derivative of a function as f'(x) is best left to good old fashioned hand-work. However, to find the derivative of a function at a point the calculator can help. This is especially useful when you need to use this "slope of the tangent" to write a tangent equation or use the rate of change in another way.
Graph the function in question and access the derivative (dy/dx) option under the calculate menu. You can then trace to the appropriate x-value or enter it as I have done in the section image. After pressing ENTER the calculator will read back the proper value. In this case the function is sin(x) whose derivative we know to be cos(x) and at the value of PI/3 we see the approximation for 0.5.7:integral (f(x)dx)
Using the definite integral as the area under the curve can be handled by the graphing menu. The example below shows the integral (f(x)dx) function used to find the area under a sine curve. Simply access the function and give the calculator the lower and upper bounds of integration when prompted. When you press ENTER it will shade in the area and give the numerical result.
The only "trick" to using this method is it works for simple definite integration. You still need to be viligent in the way the problem is worded to make sure the qusetion you are being asked is answered. (ie. when the area is also under the x-axis and therefore can been seen as negative.)