Let's start with simple substitution into an equation:
a) Find y at x=3 for y = 2x^2 + 2x + 2
The "|" is important between the equation y = 2x^2 + 2x + 2 and x=3
Let's now find x if we know y. This is a little harder as we have to solve the equation.
b) Find x at y=14 for y = 2x^2 + 2x + 2
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The easiest way to do this is to type 14 = 2x^2 + 2x + 2, highlight it using your stylus (this is important!) and then go
interactive->advanced->solve->ok
Note that it find both possible solutions (unlike using numsolve with the incorrect range specified)
Let's find the first derivative. For this use the 2D template in the soft keyboard
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c) Find the 1st derivative of y=2x^2 + 2x + 2
Note that I removed the "y=" this time. I differentiated the expression on purpose as it makes the next part easier.
Finding the 1st derivative/gradient/instantaneous rate of change at a point is also easy.
d) Find the 1st derivative of y = 2x^2 + 2x + 2 at x = 3
As you can see, it is a mix between c) and a)
Last but not least we can find a point for a particular gradient.
e) Find x at y' = 14 for y = 2x^2 + 2x + 2
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To find y itself we could repeat a)
It's very much a case of thinking what you need and then finding it. As you can see it can all be done with one line on a CAS calculator, things that would take multiple steps on paper. TanLine is also a useful function that can be investigated and used to quickly find tangents.
Viola!
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