Question

What are the coordinates of the point of intersection of the tangents when the equation of the tangent to the curve `y= x^2 -6x + 5` at the points, `x = 1` and `x = 5` is/are found?

Answer 1

The tangent equation are

`y = -4x+4` and `y = 4x-20`

Which intersect at `(3,-8)`

Explanation:

We have:

`y= x^2 -6x + 5`

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

Differentiating wrt `x` we get:

`dy/dx = 2x - 6`

So when `x=1`, we have:

# { (y= 1-6+5, = 0), (y' = 2 - 6, = -4) :} => { (((0,0))), (m=-4) :} #

So, the equation of the first tangent, Using the point/slope form `y-y_1=m(x-x_1)` is;

`y - 0 = -4 (x-1) => y = -4x+4`

And, when `x=5`, we have:

# { (y= 25-30+5, = 0), (y' = 10-6, = 4) :} => { (((5,0))), (m=4) :} #

So, the equation of the second tangent is;

`y - 0 = 4 (x-5) => y = 4x-20`

And, the point of intersection is the simultaneous solution so:

`y = -4x+4` and `y = 4x-20`

Adding gives: `2y=-16 => y=-8`
Substitution gives, `4x=12 => x=3`

Thus the coordinate of the insertion of the tangents is `(3,-8)`

We can confirm this graphically:
graph{(y-(x^2 -6x + 5))(y-(-4x+4))(y-(4x-20))=0 [-3, 8, -15, 15]}