math helppp?!?

Question

for each find the eq'n of the tangent at the given point

explain please!


y=x² @ (-2,4)


y=square root of x @ (9,3)


y=2/x+1 @91,1)

Answer 1

1) y = x^2

The first thing we need to do is find the slope of the tangent line at (-2, 4). To do this, we take the derivative.

y' = 2x

Now, we want the slope at (-2, 4), so we make x = -2. This gives us

m = 2(-2) = -4

Now that we have our slope, we use our slope formula to obtain the equation of the line.

(y2 - y1) / (x2 - x1) = m

Now, we plug in (-2, 4) for (x1, y1) and (x, y) for (x2, y2). We also know m = -4, so

(y - 4) / (x - (-2)) = -4
(y - 4) / (x + 2) = -4

y - 4 = -4(x + 2)
y - 4 = -4x - 8

So the equation of the line is

y = -4x - 4

2) y = sqrt(x) (9, 3)

y' = 1/[2sqrt(x)]

Therefore, m = 1/[2sqrt(9)] = 1/[2(3)] = 1/6

(y2 - y1) / (x2 - x1) = m
(y - 3) / (x - 9) = 1/6
y - 3 = (1/6) (x - 9)
y - 3 = (1/6)x - (9/6)
y = (1/6)x - (9/6) + 3
y = (1/6)x - (9/6) + (18/6)
y = (1/6)x + (9/6)

3) y = 2/(x + 1), at (1, 1)

y' = -2/(x + 1)^2.
Therefore

m = -2/(1 + 1)^2 = -2/(2)^2 = -2/4 = -1/2.

Therefore

(y2 - y1) / (x2 - x1) = m
(y - 1) / (x - 1) = -1/2
y - 1 = (-1/2) (x - 1)
y - 1 = (-1/2)x + 1/2
y = (-1/2)x + 3/2