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

##### 1 Answer

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:

#(y - f(x))(y - g(x))(y - h(x)) = 0#

or more generally of the form:

#f(x, y) * g(x, y) * h(x, y) = 0#

For example, we could draw

#((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#

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:

#((x-3)^100+(y-3)^100-2^100) = 0#

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

Triangles are possible too:

#((sqrt(3)/2(x-2)/(1-1/sqrt(3)(y-1)))^100+(y-sqrt(3)/2-1)^100-(sqrt(3)/2)^100) = 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

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:

#(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#

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