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:
