Let f(x)=x^2-2x+1. Find all points on the graph of f(x) whose tangent line also passes through (3,1)

Question

Let f(x) = x^2 -2x+1. Find all points on the graph of f(x) whose
tangent line also passes through the point (3,1).

I know f'(x)=2x-2

Tangent line that passes through pts.(3,1) is y=4x-11

But I don't get the part where it says to find all point on the graph??

Answer 1

Evaluate the expression for the given values. If necessary, round to the nearest tenth.(x+4y)^2 x=4 y=4

Simplify the expression by combining like terms

-9(5r+8)+3(7r+7)

Solve the equation 2(x+4)=(2x+8)=

Answer 2

 

f(x)=x²-2x+1=(x-1)², f'(x)=2x-2.

Let P(a,f(a)) be a point on the parabola, so that at that point the tangent (slope) is 2a-2.

If the line is a tangent to the curve at P, its gradient (slope) is the same as the tangent 2a-2.

Because the line passes through (3,1) the equation of the line will be y-1=(2a-2)(x-3) in slope-intercept form.

The point P must also lie on the line, so f(a)-1=(2a-2)(a-3), that is, (a-1)²-1=(2a-2)(a-3).

Therefore a²-2a=2a²-8a+6, a²-6a+6=0, a²-6a+9-9+6=0=(a-3)²-3. So a-3=±√3, a=3±√3.

Knowing a we can work out the equations of the lines:

y-1=(4+2√3)(x-3) and y-1=(4-2√3)(x-3)

Expanding:

y=(4+2√3)x-11-6√3 and y=(4-2√3)x-11+6√3.