Consider the following pair of equations:
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The value of k such that the graph of f and the graph of g only intersect one is equal to 2.
The value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.
<h3>How to find the value of the constant k of a system of two polynomic equations</h3>
Herein we have a system formed by two <em>nonlinear</em> equations, a <em>quadratic</em> equation and a <em>cubic</em> equation. Given the constraint that both function must only intersect once, we have the following expression:
f(x) - x² - k · x = 4 (1)
g(x) - x² - k · x = x³ + 2 · k (2)
x³ + 2 · k = 4
x³ + 2 · (k - 2) = 0
If f and g must intersect once, then the roots must of the form:
(x - r)³ = x³ + 2 · (k - 2)
x³ - 3 · r · x² + 3 · r² · x - r³ = x³ + 2 · (k - 2)
Then, the following conditions must be met: - 3 · r · x² = 0, 3 · r² · x = 0. If x may be any real number, then r must be zero and the value of k must be:
2 · (k - 2) = 0
k - 2 = 0
k = 2
Therefore, the value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.
To learn more on polynomic functions: brainly.com/question/24252137
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Triangle ABF has base 10 and height 3, so area (1/2)*10*3 = 15.
Trapezoid EBCD has bases 6 and 7 and height 6, so area
.. (6+7)*6/2 = 39
The total area of the figure is
.. 15 +39 = 54 . . . . . square units
Trapezoid EBCD has bases 6 and 7 and height 6, so area
.. (6+7)*6/2 = 39
The total area of the figure is
.. 15 +39 = 54 . . . . . square units