Find a point on the curve `y=x^2+x` , where the tangent is parallel to the chord joining (0, 0) and (1, 2).

Answer

`(c, f(c)) => (1/2 , 3/4)` is the point

Explanation

Let:
`f(x)=x^{2}+x`

The tangent to the curve is parallel to the chord joining the points `(0,0)` and `(1,2)`.

Assume that the chord joins the points `(a, f(a))` and `(b, f(b))`.

$$ \therefore a=0, b=1 $$
A polynomial function is continuous and differentiable at each point.

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