Product and Quotient rules

Question

I am trying to help my daughter do her college calculus homework, and it's over my head. We have worked this one for a while.

Let f(x)=(3x^2/2x+5).

Find the equation of the tangent line to f at (2,12/9)

Find the equation of the normal line to f at (2,12/9),

I appreciate the help.

Answer 1

slope of the tangent line = derivative of f(2)

for f(x) = 3x^2/(2x+5)

apply the quotient rule:

f'(x) = the denominator times the numerator, minus the numerator times the denominator, all over the denominator squared

= [(2x+5)(6x) - (3x^2)(2)]/(2x+5)^2

= (12x^2 +30x - 6x^2)/(2x+5)^2

= (6x^2 +30x)/(2x+5)^2

replace x with 2

f'(2) = (6(2)^2 +30(2))/(2(2)+5)^2

= (24 +60)/81

= 84/81= 28/27

(y-12/9) = (28/27)(x-2)

y-4/3 = (28/27)(x-2) is the tangent line's equation in point slope form

or

y = 28x/27 -56/27 +36/27

y= 28x/27 -20/27 in slope intercept form

or

28x -27y = 20 in standard form (ax+by = c form)

maybe another way to think about it is:

slope of the tangent line = 28/27 = (y-y1)/(x-x1) where (x1, y1) = a point on the line = (2, 12/9) = (2,4/3)

then the equation becomes

slope = slope=

"rise over run" = change in y over change in x

28/27 = (y -4/3)/(x-2)

cross multiply

28(x-2) = 27(y-4/3)

28x -56 = 27y - 36

28x -27y = -36+56 = 20

28x -27y = 20

the normal line has a slope which is the negative inverse of the tangent line's slope

take 28/27, flip it upside down and change its sign to get -27/28 = slope of the normal line

"normal" means perpendicular

-27/28 = (y- 4/3)/(x-2)

or

y-4/3 = (-27/28)(x-2) is the normal line's equation in point slope form

or

multiply by 28 to eliminate the fraction

28y -4(28)/3 = -27x +54

27x +28y = 54 + 132/3

27x +28y = (162+132)/3 = 294/3

or

81x + 84y = 294

(these calculations get a little tedious. No guarantees there's no arithmetic mistakes along the way. But this is the basic method

find two expressions for slope, set them equal

that gives you the equation of the tangent or normal line)

You might also go to a graphing calculator, either online or handheld and graph the function, as it's not obvious what it is. Desmos had a good online graphing calulator. Then you can see what the slope is likely to be at that point.

in some of these problems there's 0ccasionally a misprint or miscopied problem, so it's worth checking to see if the given point is really on the curve of the function. This time it is, but you can waste hours trying to solve it if it's not.