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Sagot :
To establish a quadratic function that defines all the points (x, y) of the cable, we can use the vertex form of a quadratic equation:
\[ y = a(x - h)^2 + k \]
Where:
- \( h \) and \( k \) are the coordinates of the vertex (the lowest point of the cable in this case).
- \( a \) is a constant that determines the width of the parabola.
Given that the maximum sag of the cable is 10m and it occurs midway between the support towers, the vertex of the parabola is at the point (40, -10). Since the towers are 80m apart, the cable is symmetric, and its vertex is the midpoint.
Now, we can plug the vertex coordinates into the equation and solve for \( a \):
\[ y = a(x - 40)^2 - 10 \]
\[ -10 = a(0 - 40)^2 - 10 \]
\[ -10 = 1600a - 10 \]
\[ 0 = 1600a \]
\[ a = 0 \]
Since \( a = 0 \), the equation simplifies to:
\[ y = -10 \]
So, the equation of the cable is \( y = -10 \), which means the cable is a straight line parallel to the x-axis and located 10m below it.
\[ y = a(x - h)^2 + k \]
Where:
- \( h \) and \( k \) are the coordinates of the vertex (the lowest point of the cable in this case).
- \( a \) is a constant that determines the width of the parabola.
Given that the maximum sag of the cable is 10m and it occurs midway between the support towers, the vertex of the parabola is at the point (40, -10). Since the towers are 80m apart, the cable is symmetric, and its vertex is the midpoint.
Now, we can plug the vertex coordinates into the equation and solve for \( a \):
\[ y = a(x - 40)^2 - 10 \]
\[ -10 = a(0 - 40)^2 - 10 \]
\[ -10 = 1600a - 10 \]
\[ 0 = 1600a \]
\[ a = 0 \]
Since \( a = 0 \), the equation simplifies to:
\[ y = -10 \]
So, the equation of the cable is \( y = -10 \), which means the cable is a straight line parallel to the x-axis and located 10m below it.
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