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1. Write an equation, in point-slope form, of the line with a slope of 2 and containing the point (5, -2).

A.

y + 2 = 2(x - 5)

B.

y - 2 = 2(x + 5)

C.

y - 5 = 2(x + 2)

2. Find the equation of a vertical line through the point (-1, 6).

A.

x = 6

B.

x = -1

C.

y = 6

D.

y = -1

3. Find the equation of a line through the points (1, 3) and (6, -7). Express the equation in slope-intercept form.

A.

y = 2x + 1

B.

y = -2x + 5

C.

y = -2x + 7

4. Find the equation of the line that is perpendicular to y = 5 that goes through the point (-3, 2).

A.

y = -1

B.

y = 2

C.

x = 5

D.

x = -3

5. Write the equation of the line with a slope of -2 and goes through the point (-1, 6). Express the equation in standard form.

A.

2x + y = -7

B.

2x + y = -11

C.

2x + y = 4

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The slope of the tangent is where the derivative is 0. So you have to use product rule and implicit differentiation to get:

3x^2 + 3y^2 * y' = 3xy' + 3y

Solving for y':
3y^2 y' - 3xy' = 3y - 3x^2
y' [3y^2 - 3x] = 3y - 3x^2
y' = (3y - 3x^2) / (3y^2 - 3x)
y' = (y - x^2) / (y^2 - x)

When y' = 0 gives 0 = y - x^2 => At (x, y) the gradient is m = x^2/y

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the slope formula is = (Y1 - Y2)/(X1 - X2)

(-1,4) = (X1, Y1)
(5,-8) = (X2, Y2)

(4 -(-8)/(-1 - 5)
(4 + 8)/-6
12/-6
-2

slope = -2

Hope this Helps!

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