Question

A tangent line and its normal line have a point of tangency to the function f(x) at (x,y). If the slop of the normal line is m=(8/7), what are the coordinates of the point of tangency?

Please help, step by step explanation would be great

Answer 1

it's 16,-8

Explanation:

other dude is way off, by the way, 49/900 isn't approximatley .544, its .054

Answer 2

`\displaystyle \\\left((49)/(900),(3761)/(900)\right)` or approximately (0.544, 4.179)

Explanation:

A function and its tangent lines intersect when their slopes are the same. Find the x-coordinate when the slope of f(x) is equal to 8/7 by taking the derivative of f(x):

`\displaystyle\\f(x)=√(x)-x+4\\f'(x)=(1)/(2)\cdot(1)/(√(x))-1=(1)/(2√(x))-1`

Set
`f'(x)` equal to 8/7 and solve for x:

`\displaystyle \\(1)/(2√(x))-1=(8)/(7),\\x=(49)/(900)`

Therefore,
`f(x)` will intersect at a point of tangency with a line of slope 8/7 at x=49/900. Plug in x=49/900 into
`f(x)` to get the y-coordinate:

`\displaystyle\\y=√(x)-x+4 \vert_(x=49/900)=(3761)/(900)`

⇒Answer: (49/900, 3761/900) or approximately (0.544, 4.179)