Find the equations of the tangents to y=x^2 which pass through the point (2,3).

Mathematics

1 answer:

algol [13]2 years ago

6 0

Answer: The equations of the tangents that pass through

y = -x- 1 and

y = 11x -25

Step-by-step explanation: The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have

y= x^2 +x

Differentiating wrt x we get:

dy/dx = 2x +1

Let P (a, B) be any generic point on the curve. Then the gradient of the tangent at P is given by:

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