@Thalassokrator said in #29:
> Yes, we could add m. The y-intercept would then be m + 1.
> cosh(x) crosses the y-axis at (x, y) = (0, 1) from the get go:
>
> y = cosh(x) = (e^x + e^(-x))/2 so plugging in x = 0 and e^0 = 1 yields
> y(0) = cosh(0) = (1 + 1/1)/2 = 2/2 = 1
>
> Adding a variable m only shifts the function up or down (depending on the sign of m) but doesn't change the shape into a parabola. Adding m only changes the constant term in the resulting function's Taylor series, it doesn't get rid of any higher order terms.
m would be m + 1. However, the question is, y = cosh(x) + m is clearly a parabola.
> Yes, we could add m. The y-intercept would then be m + 1.
> cosh(x) crosses the y-axis at (x, y) = (0, 1) from the get go:
>
> y = cosh(x) = (e^x + e^(-x))/2 so plugging in x = 0 and e^0 = 1 yields
> y(0) = cosh(0) = (1 + 1/1)/2 = 2/2 = 1
>
> Adding a variable m only shifts the function up or down (depending on the sign of m) but doesn't change the shape into a parabola. Adding m only changes the constant term in the resulting function's Taylor series, it doesn't get rid of any higher order terms.
m would be m + 1. However, the question is, y = cosh(x) + m is clearly a parabola.