Non homogeneous recurrence relation2fn - 6fn-1 = (n+1)(3^n)

Mathematics

1 answer:

mel-nik [20]2 years ago

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Answer:

Non-Homogeneous Recurrence Relation and Particular Solutions

A recurrence relation is called non-homogeneous if it is in the form. Fn=AFn−1+BFn−2+f(n) where f(n)≠0.

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The equation of a quadratic is =2^2+3−2and after you factorizing 2^2+3−2=0you got x = -2 and x = 1/2 and the curve crosses the y

Lelechka [254]

Given the next quadratic function:

$y=2x^2+3x-2$

to sketch its graph, first, we need to find its vertex. The x-coordinate of the vertex is found as follows:

$x_V=\frac{-b}{2a}$

where <em>a</em> and <em>b</em> are the first two coefficients of the quadratic function. Substituting with a = 2 and b = 3, we get:

$\begin{gathered} x_V=\frac{-3}{2\cdot2} \\ x_V=-\frac{3}{4}=-0.75 \end{gathered}$

The y-coordinate of the vertex is found by substituting the x-coordinate in the quadratic function, as follows:

$\begin{gathered} y_V=2x^2_V+3x_V-2 \\ y_V=2\cdot(-\frac{3}{4})^2+3\cdot(-\frac{3}{4})-2 \\ y_V=2\cdot\frac{9}{16}+3\cdot(-\frac{3}{4})-2 \\ y_V=\frac{9}{8}-\frac{9}{4}-2 \\ y_V=-\frac{25}{8}=-3.125 \end{gathered}$

The factorization indicates that the curve crosses the x-axis at the points (-2, 0) and (1/2, 0). We also know that the curve crosses the y-axis at (0,-2). Connecting these points and the vertex (-0.75, -3.125) with a U-shaped curve, we get: