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?

1 Answer
Apr 28, 2018

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]}