# Can we make diagrams for an answer or is it necessary to upload a picture?

Nov 30, 2016

See explanation...

#### Explanation:

In a limited way, it is possible to construct some simple diagrams using the graphing capability, but generally it is easier to construct them externally using a vector graphic editor like the (free) Inkscape or GIMP image editing applications.

You can plot more than one curve at a time by plotting an equation of the form:

$\left(y - f \left(x\right)\right) \left(y - g \left(x\right)\right) \left(y - h \left(x\right)\right) = 0$

or more generally of the form:

$f \left(x , y\right) \cdot g \left(x , y\right) \cdot h \left(x , y\right) = 0$

For example, we could draw $3$ circles by graphing:

$\left({\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} - 0.2\right) \left({\left(x - 2\right)}^{2} + {\left(y - 1\right)}^{2} - 0.2\right) \left({\left(x - \frac{3}{2}\right)}^{2} + {\left(y - 2\right)}^{2} - 0.2\right) = 0$

graph{((x-1)^2+(y-1)^2-0.2)((x-2)^2+(y-1)^2-0.2)((x-3/2)^2+(y-2)^2-0.2) = 0 [-4.54, 5.46, -1.46, 3.54]}

Small circles can be used to indicate points, e.g.:

graph{(y - 1/4x^2)((x-2)^2+(y-1)^2-0.02)((x-4)^2+(y-4)^2-0.02) = 0 [-8.62, 11.38, -2.96, 7.04]}

Squares (or at least a good approximation) can be drawn by plotting something like:

$\left({\left(x - 3\right)}^{100} + {\left(y - 3\right)}^{100} - {2}^{100}\right) = 0$

graph{((x-3)^100+(y-3)^100-2^100) = 0 [-8.62, 11.38, -2.96, 7.04]}

Triangles are possible too:

$\left({\left(\frac{\sqrt{3}}{2} \frac{x - 2}{1 - \frac{1}{\sqrt{3}} \left(y - 1\right)}\right)}^{100} + {\left(y - \frac{\sqrt{3}}{2} - 1\right)}^{100} - {\left(\frac{\sqrt{3}}{2}\right)}^{100}\right) = 0$

graph{((sqrt(3)/2(x-2)/(1-1/sqrt(3)(y-1)))^100+(y-sqrt(3)/2-1)^100-(sqrt(3)/2)^100) = 0 [-3.58, 6.42, -1.26, 3.74]}

It is possible to limit the range of a curve by using square roots, since the domain of $\sqrt{\ldots}$ is limited to non-negative values.

For example, here's a simple diagram of a parabolic trajectory over a candle...

graph{(y-sqrt(x(2.2-x)*0.3188/1.21)^2)((5(x-1.1))^50+(y-0.125)^50-0.125^50)((2(x-1.1))^2+(y-0.27)^2-0.02^2) = 0 [-0.191, 2.309, -0.565, 0.685]}

This is generated by graphing:

$\left(y - {\sqrt{x \left(2.2 - x\right) \cdot \frac{0.3188}{1.21}}}^{2}\right) \left({\left(5 \left(x - 1.1\right)\right)}^{50} + {\left(y - 0.125\right)}^{50} - {0.125}^{50}\right) \left({\left(2 \left(x - 1.1\right)\right)}^{2} + {\left(y - 0.27\right)}^{2} - {0.02}^{2}\right) = 0$

The plot of the parabola is limited to non-negative values by taking its square root and squaring the result.