Two tangents to the curve f (x) = x² - 8x + 4 intersect at the point (5, -20). A straight line runs through the two tangent points.

Question

Determine the coordinates of the intersection of this line with the x-axis.

Please help me walk through this step by step.

Answer 1

A tangent line at x=a is y = f(a) + f'(a) (x-a)

f'(x) = 2x-8

So a tangent line at x=a will be y = a² - 8a + 4 + (2a-8) (x-a)

When x= 5 the value of this must be - 20 (as the tangent lines pass through (5, -20)).

So a² - 8a + 4 + (2a-8) (5-a) = -20

a² - 8a + 4 + 10a -2a² -40 + 8a = - 20

-a² +10a -36 = - 20

-a² + 10a -16 = 0

a² - 10a + 16 = 0

(a -2) (a-8) = 0

a = 2, a= 8

So the tangents will be when x=2, x=8

x = 2, f(2) = 4 - 16 + 4 = -8

x = 8, f(8) = 64 - 64 + 4 = 4

So the two intersection points of tangents with function are (2,-8) and (8, 4)

Line through these points (y--8)/ (x-2) = (4--8) / (8-2)

(y+8)/(x-2) = 12/6 = 2

(y+8)/(x-2) = 2

y+8 = 2x-4

y = 2x-12

On x-axis y=0

2x-12 = 0

x = 6

So required point is (6,0)