Yes. A monomial is a term composed by a coefficient and powers of variables multiplied with each other, for example
is a monomial.
In your case, the monomials are
As the name suggests, a trinomial is the sum of three monomials, which is exactly your case.
(3-8y) x (-2.5)
= -7.5+20y
![\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 3}}\quad ,&{{ 1}})\quad \end{array}\\\\ -----------------------------\\\\ 2\left[ x^4+2x^2y^2+y^4 \right]=25(x^2-y^2)\qquad thus \\\\\\ 2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[2x-2y\frac{dy}{dx} \right] \\\\\\ 2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=50\left[x-y\frac{dy}{dx} \right] \\\\\\ ](https://tex.z-dn.net/?f=%5Cbf%202%5Bx%5E2%2By%5E2%5D%5E2%3D25%28x%5E2-y%5E2%29%5Cqquad%20%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%5C%5C%0A%25%20%20%20%28a%2Cb%29%0A%26%28%7B%7B%203%7D%7D%5Cquad%20%2C%26%7B%7B%201%7D%7D%29%5Cquad%20%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A2%5Cleft%5B%20x%5E4%2B2x%5E2y%5E2%2By%5E4%20%5Cright%5D%3D25%28x%5E2-y%5E2%29%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0A2%5Cleft%5B%204x%5E3%2B2%5Cleft%5B%202xy%5E2%2Bx%5E22y%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%2B4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%3D25%5Cleft%5B2x-2y%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A2%5Cleft%5B%204x%5E3%2B2%5Cleft%5B%202xy%5E2%2Bx%5E22y%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%2B4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%3D50%5Cleft%5Bx-y%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A)
![\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx} \right] \\\\\\ 4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x \\\\\\ \cfrac{dy}{dx}[4x^2y+4y^3+25y]=25x-4x^3+4xy^2 \\\\\\ \cfrac{dy}{dx}=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}\impliedby m=slope](https://tex.z-dn.net/?f=%5Cbf%20%5Cleft%5B%204x%5E3%2B2%5Cleft%5B%202xy%5E2%2Bx%5E22y%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%2B4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cright%5D%3D25%5Cleft%5Bx-y%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C%0A4x%5E3%2B4xy%5E2%2B4x%5E2y%5Cfrac%7Bdy%7D%7Bdx%7D%2B4y%5E3%5Cfrac%7Bdy%7D%7Bdx%7D%2B25y%5Cfrac%7Bdy%7D%7Bdx%7D%3D25x%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%5B4x%5E2y%2B4y%5E3%2B25y%5D%3D25x-4x%5E3%2B4xy%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B25x-4x%5E3%2B4xy%5E2%7D%7B4x%5E2y%2B4y%5E3%2B25y%7D%5Cimpliedby%20m%3Dslope)
notice... a derivative is just the function for the slope
now, you're given the point 3,1, namely x = 3 and y = 1
to find the "m" or slope, use that derivative, namely
that'd give you a value for the slope
to get the tangent line at that point, simply plug in the provided values
in the point-slope form
and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1
