How to find the equation of this tangent?

[Jeanclaud]

Mod note: Thread moved from Precalc section

Homework Statement

F(x)=sqrt(-2x^2 +2x+4)
1.discuss variation of f and draw (c)
2.find the equation of tangent line to (c) that passes through point A(-2,0)

The Attempt at a Solution

I solved first part I found the domain of definition and f'(x) and I drew (c) but in the second part A does not belong to the domain of definition so i can't use f'(x) what should i do? Please help.

 

[haruspex]

I solved first part I found the domain of definition and f'(x) and I drew (c) but in the second part A does not belong to the domain of definition so i can't use f'(x) what should i do? Please help.

What is the equation of a line that is tangent to the curve at x?

 

[Jeanclaud]

What is the equation of a line that is tangent to the curve at x?

Y=f'(a)(x-a)+f(a) at point of abscissa a

 

[haruspex]

Y=f'(a)(x-a)+f(a) at point of abscissa a

Right. So substitute for f and f' in there. Plug in the fact that this tangent is to pass through A. What equation does that give you?

 

[HallsofIvy]

haruspex, that is not Jeanclaude's point. -2x^2+ 2x+ 4= -2(x+ 1)(x- 2) is positive only for x between -1 and 2. It is negative for al x< -1 and so \sqrt{-2x^2+ 2x+ 4} does NOT EXIST for x< -1 and, in particular, at x= -2.

Jeanclaude, the fact that -2 is not in the domain means there is NO function value at x= -2 so no graph there and NO tangent line

 

[SteamKing]

The wording of this problem is a little tricky.

1. f(x) certainly doesn't exist at point A.
2. The problem statement asks for the equation of a line tangent to f(x) which passes thru point A. It does not specify that the line is tangent to f(x) at point A, merely that the tangent line must pass thru this point and be tangent to f(x) at some other unspecified point.

 

[HallsofIvy]

Oh, right, I misread the problem. Thanks, SteamKing.

The derivative of F is \frac{1}{2}(-2x^2+ 2x+ 4)^{-1/2}(-4x+ 2). The equation of the tangent line at any point (x_0, \sqrt{-x_0^2+ 2x_0+ 4}) is y= \frac{1}{2}(-2x_0^2+ 2x_0+ 4)^{-1/2}(-4x_0+ 2)(x- x_0)+ \sqrt{-x_0^2+ 2x_0+ 4}). Saying that goes through (-2, 0) means that we must have
0= \frac{1}{2}(-2x_0^2+ 2x_0+ 4)^{-1/2}(-4x_0+ 2)(-2- x_0)+ \sqrt{-x_0^2+ 2x_0+ 4})
Solve that equation for x_0.

 

[haruspex]

haruspex, that is not Jeanclaude's point. -2x^2+ 2x+ 4= -2(x+ 1)(x- 2) is positive only for x between -1 and 2. It is negative for al x< -1 and so \sqrt{-2x^2+ 2x+ 4} does NOT EXIST for x< -1 and, in particular, at x= -2.

Jeanclaude, the fact that -2 is not in the domain means there is NO function value at x= -2 so no graph there and NO tangent line

Funny, I thought perhaps JC had misread the question that way, but then just decided he didn't know how to find a tangent through a point that is not on the curve. Thank you SteamKing for covering the misreading option.

 

[Ray Vickson]

Mod note: Thread moved from Precalc section

Homework Statement

F(x)=sqrt(-2x^2 +2x+4)
1.discuss variation of f and draw (c)
2.find the equation of tangent line to (c) that passes through point A(-2,0)

The Attempt at a Solution

I solved first part I found the domain of definition and f'(x) and I drew (c) but in the second part A does not belong to the domain of definition so i can't use f'(x) what should i do? Please help.

The point A is on a tangent line to the curve $y = F(x)$; it need not be in the domain of definition of $F$. In other words, nobody is saying that we want the tangent at the point $x = -2$; they are just saying that the line which is tangent to the curve at some point should also pass through the point A = (-2,0).