Equations: x = 2 -2y and x^2 +4y^2 = 4
If: x = 2 -2y
Then: x^2 = 4y^2 -8y +4
If: x^2 +4y^2 = 4
Then: 4y^2 -8y +4 +4y^2 = 4
Collecting and transposing terms: 8y^2 -8y = 0 or y^2 -y = 0
Factorizing the above: (y-1)(y+0) = 0 meaning y = 1 or 0
By substitution points of contact are made at: (0, 1) and (2, 0)
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
The tangent line is the instantaneous rate of change at a point on a curve. The secant line crosses a curve twice at points A and B, representing the average rate of change between those two points.
The solutions to a quadratic equation on a graph are the two points that cross the x-axis. NB A graphed quadratic equ'n produces a parabolic curve. If the curve crosses the x-axis in two different points it has two solution. If the quadratic curve just touches the x-axis , there is only ONE solution. It the quadratic curve does NOT touch the x-axis , then there are NO solutions. NNB In a quadratic equation, if the 'x^(2)' value is positive, then it produces a 'bowl' shaped curve. Conversely, if the 'x^(2)' value is negative, then it produces a 'umbrella' shaped curve.
The two solutions are (x, y) = (-0.5, -sqrt(3.5)) and (-0.5, sqrt(3.5))
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If: y = x2-x-12 Then points of contact are at: (0, -12), (4, 0) and (-3, 0)
It works out that line 3x-y = 5 makes contact with the curve 2x^2 +y^2 = 129 at (52/11, 101/11) and (-2, -11)
If 2x+5y = 4 and y^2 = x+4 then by combining the equations into a single quadratic equation and solving it the points of contact are made at (12, -4) and (-7/2, 3/2)
The tangent line is the instantaneous rate of change at a point on a curve. The secant line crosses a curve twice at points A and B, representing the average rate of change between those two points.
Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)
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The solutions to a quadratic equation on a graph are the two points that cross the x-axis. NB A graphed quadratic equ'n produces a parabolic curve. If the curve crosses the x-axis in two different points it has two solution. If the quadratic curve just touches the x-axis , there is only ONE solution. It the quadratic curve does NOT touch the x-axis , then there are NO solutions. NNB In a quadratic equation, if the 'x^(2)' value is positive, then it produces a 'bowl' shaped curve. Conversely, if the 'x^(2)' value is negative, then it produces a 'umbrella' shaped curve.
It is (-0.3, 0.1)
Find the roots of the equation that defines the curve. these are the points where the curve crosses the x-axis.Then integrate between the first end-point and first root, first and second root, etc until the last root and the second end-point. Add together the absolute values of the integrals.Find the roots of the equation that defines the curve. these are the points where the curve crosses the x-axis.Then integrate between the first end-point and first root, first and second root, etc until the last root and the second end-point. Add together the absolute values of the integrals.Find the roots of the equation that defines the curve. these are the points where the curve crosses the x-axis.Then integrate between the first end-point and first root, first and second root, etc until the last root and the second end-point. Add together the absolute values of the integrals.Find the roots of the equation that defines the curve. these are the points where the curve crosses the x-axis.Then integrate between the first end-point and first root, first and second root, etc until the last root and the second end-point. Add together the absolute values of the integrals.
Points on the Curve was created on 1984-01-16.