Answer:
I attached a picture of the completed picture below.
The values given to you in the picture are shown in pink
Use those values to fill in the rest, which are shown in blue.
Find the area of polygon
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Answer:
(a) The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)
(b) The gradient of the curve at each point of intersection are;
Gradient at (-2, -12/5) = -0.92
Gradient at (2/5, 0) = 4.3
Gradient at (2, 8/5) = -0.28
Step-by-step explanation:
The equations of the lines are;
5·x - 5·y = 2......(1)
x²·y - 5·x + y + 2 = 0.......(2)
Making y the subject of equation (1) gives;
5·y = 5·x - 2
y = (5·x - 2)/5
Making y the subject of equation (2) gives;
y·(x² + 1) - 5·x + 2 = 0
y = (5·x - 2)/(x² + 1)
Therefore, at the point the two lines intersect their coordinates are equal thus we have;
y = (5·x - 2)/5 = y = (5·x - 2)/(x² + 1)
Which gives;
Therefore, 5 = x² + 1
x² = 5 - 1 = 4
x = √4 = 2
Which is an indication that the x-coordinate is equal to 2
The y-coordinate is therefore;
y = (5·x - 2)/5 = (5 × 2 - 2)/5 = 8/5
The coordinates of the points of intersection = (2, 8/5}
Cross multiplying the following equation
Substituting the value for y in equation (2) with (5·x - 2)/5 gives;
Therefore;
5·x³ - 2·x² - 20·x + 8 = 0
(x - 2)×(5·x² - b·x + c) = 5·x³ - 2·x² - 20·x + 8
Therefore, we have;
x²·b - 2·x·b -x·c + 2·c -5·x³ + 10·x²
5·x³ - 10·x² - x²·b + 2·x·b + x·c - 2·c = 5·x³ - 2·x² - 20·x + 8
∴ c = 8/(-2) = -4
2·b + c = - 20
b = -16/2 = -8
Therefore;
(x - 2)×(5·x² - b·x + c) = (x - 2)×(5·x² + 8·x - 4)
(x - 2)×(5·x² + 8·x - 4) = 0
5·x² + 8·x - 4 = 0
x² + 8/5·x - 4/5 = 0
(x + 4/5)² - (4/5)² - 4/5 = 0
(x + 4/5)² = 36/25
x + 4/5 = ±6/5
x = 6/5 - 4/5 = 2/5 or -6/5 - 4/5 = -2
Hence the three x-coordinates are
x = 2, x = - 2, and x = 2/5
The y-coordinates are derived from y = (5·x - 2)/5 as y = 8/5, y = -12/5, and y = y = 0
The coordinates of the points of intersection are (-2, -12/5), (2/5, 0), and (2, 8/5)
(b) The gradient of the curve, , is given by the differentiation of the equation of the curve, x²·y - 5·x + y + 2 = 0 which is the same as y = (5·x - 2)/(x² + 1)
Therefore, we have;
.......(3)
Which gives by plugging in the value of x in the slope equation;
At x = -2, = -0.92
At x = 2/5, = 4.3
At x = 2, = -0.28
Therefore;
Gradient at (-2, -12/5) = -0.92
Gradient at (2/5, 0) = 4.3
Gradient at (2, 8/5) = -0.28.