Find the gradients of the lines that pass through the point (1,7) and are tangent to the parabola y = (2 - x) (1 + 3x).

Question

Working out please!

Atep by step..

Thank you!

Answer 1

The equation of the tangent line is (Temporarily) y = mx + (7-m)

They must meet at exactly one point:

mx + 7 - m = (2-x)(1+3x)

mx + 7 - m = 2 + 6x - x - 3x^2

mx + 7 - m = 2 + 5x - 3x^2

0 = -3x^2 + (5-m)x + 2- 7 + m

0 = -3x^2 + (5-m)x + (m-5)

the discriminant must be zero

0 = B^2 - 4AC = (5-m)^2 - 4(-3)(m-5)

= 25 - 10m + m^2 - (-12)(m-5)

= 25 - 10m + m^2 + 12(m-5)

= 25 - 10m + m^2 + 12m - 60

= m^2 + 2m - 35

= ( m + 7 )( m - 5 )

m = -7 or m = 5

y = -7x + 15

y = 5x + 2 <--- this one works !!!!