Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$; normal $y = x - 1$
how could this be?

Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$; normal $y = x - 1$
how could this be?
Thus,
$y'=-\frac{8}{x^3}$
$x=2$
Thus, (the derivative at that point),
$y'(2)=-\frac{8}{8}=-1$
Using slope-intercept formula,
$y-y_0=-1(x-x_0)$
$(x_0,y_0)=(2,1)$
Thus,
$y-1=-x-2$
Thus,
$y=-x-1$
This is equation of tangent line.

Thus you are correct in saying the book is wrong.

Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

$y = \frac{4}{x^2};(2,1)$

the answer from the back of my book is:
tangent: $y=-x+3$; normal $y = x - 1$
how could this be?
The tangent line will be of the form $y = mx + b$.

We can get the slope from the derivative:
$y' = -\frac{8}{x^3}$

At x = 2 this becomes: $y' = -\frac{8}{8} = -1$

So we need the intercept. At x = 2 we know that y = 1, so
$1 = -1 \cdot 2 + b$

$b = 3$

So the tangent line is $y = -x + 3$

The line normal to this will have a slope of $-\frac{1}{-1} = 1$, so the form of the normal line will be $y = x + b'$ and passes through (2, 1), so
$1 = 2 + b'$

$b' = -1$

So the normal line is $y = x - 1$.

(Sorry about the graph. The lines don't look perpendicular because the x and y scales are different. I haven't figured out how to fix this yet.)

-Dan