Find a point on the curve y = 3x² – 9x + 8 at which the tangents are equally inclined with the axes.

Given:

The curve is y = 3x² – 9x + 8

Differentiating the above w.r.t x

⇒ y = 3x² – 9x + 8

⇒ = 23x^{2 – 1} – 9 + 0

⇒ = 6x – 9 ...(1)

Since, the tangent are equally inclined with axes

i.e, or

= The Slope of the tangent = tan

⇒ = tan() or tan()

⇒ = 1or – 1 ...(2)

tan() = 1

From (1) & (2),we get,

⇒ 6x – 9 = 1 0r 6x – 9 = – 1

⇒ 6x = 10 0r 6x = 8

⇒ x = or x =

⇒ x = or x =

Substituting x = or x = in y = 3x² – 9x + 8,we get,

When x =

⇒ y = 3()² – 9() + 8

⇒ y = 3() – () + 8

⇒ y = () – () + 8

taking LCM = 9

⇒ y = ()

⇒ y = ()

⇒ y = ()

⇒ y = ()

when x =

⇒ y = 3()² – 9() + 8

⇒ y = 3() – () + 8

⇒ y = () – () + 8

taking LCM = 9

⇒ y = ()

⇒ y = ()

⇒ y = ()

⇒ y = ()

Thus, the required point is (,) & (,)