Find equation of tangent line of the function f(x)= (x-7)/(x-1) at (3,-1)

Question

Find the equation of the tangent line of the rational function, f(x)= (x-7)/(x-1) at the point (3,-1). Keep as point-slope form.

Answer 1

y = (x-7)/(x-1) = (x-7)(x-1)^-1

y'= (x-1) - (x-7))/(x-1)^2 = 6/(x-1)^2

y'(3) = 6/(3-1)^2 = 6/4 = 3/2

y'(3) = 3/2 = slope of the tangent line at (3,-2) y(3)= (3-7)/(3-1) = -4/2 = -2

tangent's equation is y+2 = (3/2)(x-3) in point slope form

or

y = 1.5x -6.5 in slope intercept form

Answer 2

The point of tangency to:

f(x)= (x-7)/(x-1)

at (3,-1) has been corrected to (3, -2), which lies on the curve y = f(x)… (3-7)/(3-1) = -4/2 = -2

With something like f(x)= (x-7)/(x-1)… the usual procedure for the derivative would be the ‘quotient’ rule… perfectly valid.

There is a completely optional shortcut here: f(x) = (x - 1 - 6)/(x - 1) = 1 - 6/(x - 1) ==>

Given that differentiating 1/(x - 1) will be almost identical to differentiating 1/x (which produces -1/x² ), we quickly see that:

f’(x) = (-6)(-1)/(x - 1)² = 6 / (x - 1)²

f’(3) = 6/2² = 3/2

Point-Slope form: y - y₁ = m(x - x₁) [**]

(y - (-2)) = (3/2)(x - 3) ==> y + 2 = (3/2)(x - 3)

Slope-Intercept form:

y + 2 = (3/2)x - 9/2 ==> y = (3/2)x - 9/2 - 4/2 ==> y = (3/2)x - 13/2

[**] For jotting something down quickly, I often like to use a modified Point-slope form (which you will sometimes see in Calculus books):

(y - y₁) / (x - x₁) = m

which may be intuitive, if only because slope ‘m’, after all, means “rise” / “run”.

Of course, if you are asked for the equation of a line in Point-slope form, the expectation is the traditional form.

Answer 3

First make sure the point (3, -2) is on the given function:

f(3) = (3-7)/(3-1) = -4/2 = -2 ; check

Use the quotient rule to determine the derivative. Note that the rational function has a vertical asymptote at x = 1, so the derivative will not be defined at x = 1 either.

f'(x) = [(x-1)1 - (x-1)1]/(x-1)²= 6/(x-1)²

The value of the derivative at x = 3:

f'(3) = 6/(3-1)²= 6/4 = 3/2

So the equation of the tangent line using point slope:

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

y = (3/2)x - 13/2

Visit this graph and use the moveable point to see different tangent lines along with corresponding slope. The light green dashed graph is the graph of the derivative and it is also evaluated at the x-coordinate of the moveable point to show that its y-coordinate is the slope of the tangent.

desmos.com/calculator/b6qfneslsu

Answer 4

Hope this makes sense!!! :)

Answer 5

Once you have the slope = 3/2, use y=mx+b with the slope and a point to find the b.

-2 = (3/2)^3 +b

-2 = 9/2 + b

-2 - 9/2 = -4/2 -9/2 = -13/2 = b.

The equation of the tangent line is y = (3/2)x - 13/2