# functions / graphs Given the point (3,0) and the tangent line passing through the point, what is the point on the graph of y=1+(1/x) to acomplish the condition?

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BY definition, the tangent line to a graph, at a given point, is the derivative of the given function:

dy/dx = -1/x^2

The slope of the tangent line is:

m = -1/3^2

m = -1/9

The equation of the line that passes through the point (3,0) and it has the slope m = -1/9 is:

y - 0 = (-1/9)(x - 3)

y = -x/9 + 1/3

On the other hand, the tangent line is passing through the point (3, 0) and the slope of a line that passes through 2 points is:

m = (y - 0)/(x - 3)

m = y/(x - 3)

We'll put m = dy/dx:

y/(x - 3) = -1/x^2

We'll cross multiply and we'll get:

x^2*y = 3 - x

x^2(1 + 1/x) = 3 - x => x^2 + x + x - 3 = 0 => x^2 + 2x - 3 = 0

The quadratic could be determined knowing the values of sum and the product of the roots:

x^2 - Sx + P = 0

S = -2

P = -3

x1 = -3 and x2 = 1

-3+1 = -2 = S

-3*1 = -3 = P

The quadratic could be written as a product of linear factors:

(x-1)(x+3) = 0

x1 - 1 = 0

x1 = 1

x2 + 3 = 0

x2 = -3

y1 = 1 + 1/x1

y1 = 1 + 1

y1 = 2

y2 = 1 - 1/3

y2 = 2/3