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Q: How do you find the exact value of the gradient of the curve with equation y equals 1 divide by 4 minus x2 at the point where x equals 1 over 2?

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The line described by the equation is a hyperbola and its gradient is different along its length. Its gradient is -10/(5x-3)2 except where x = 3/5 and the curve is not defined. The general form of a parallel curve is y = a + 2/[5(x + b) - 3] where a and b are constants.

A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.

Cosine

Not a constant, but the differential, i.e. gradient, of the equation. It = 0 at maxima and minima, where the curve is at its turning-point(s).

Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across

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Gradient to the curve at any point is the derivative of y = x2 So the gradient is d/dx of x2 = 2x. When x = 2, 2x = 4 so the gradient of the tangent at x = 2 is 4.

The line described by the equation is a hyperbola and its gradient is different along its length. Its gradient is -10/(5x-3)2 except where x = 3/5 and the curve is not defined. The general form of a parallel curve is y = a + 2/[5(x + b) - 3] where a and b are constants.

A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.

If the gradient is a positive number the curve is increasing, and if the gradient is a negative number it is decreasing.

Cosine

y=0. note. this is a very strange "curve". If y=0 then any value of x satisfies the equation, leading to a curve straight along the y axis. For any non-zero value of y the curve simplifies to y = -x. The curve is not differentiable at the origin.

The gradient of the tangents to the curve.

Differentiate the curve twice and then enter a value for x. If the answer is positive, the gradient is increasing at that point. If the answer is negative, the gradient is decreasing at that point. And if the answer is zero, the gradient is not changing.

Not a constant, but the differential, i.e. gradient, of the equation. It = 0 at maxima and minima, where the curve is at its turning-point(s).

The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.

y2=x3+3x2

Draw a tangent to the curve at the point where you need the gradient and find the gradient of the line by using gradient = up divided by across

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