Equation of Tangent Line

What is the equation of the tangent to the curve f(x)= 4x²+2x-1 at the point x=0?

• y+1=2(x+1)

• y = 2x - 1

• y=2x+2

• y = 2x -3

f’(x) = 3x+4. Find the slope of the line tangent to f(x) at the point (0, 7)

• 3

• 4

• -7

• 7

Find the equation of the tangent line of the function:
x³ - 7xy + y³ = -125 at (2,1)

• y - 1 = 0.45 (x - 2)

• y - 1 = -0.45 (x - 2)

• y - 1 = 2.2 (x - 2)

• y - 1 = -2.2 (x - 2)

Find the equation of the tangent line of the function:
(y - 5)⁵ = x² + 2xy - 33 at (-3, -1)

• y + 1 = -0.006 (x + 3)

• y - 1 = -0.006 (x - 3)

• y + 1 = 160.75 (x + 3)

• y - 1 = 160.75 (x - 3)

• y - 1 = 0.6 (x - 1)
• y - 1 = -0.6 (x - 1)
• y - 1 = 1.67 (x - 1)
• y - 1 = -1.67 (x - 1)

• y - 2 = 0.93 (x - 3)
• y - 2 = -15/14 (x - 3)
• y - 2 = 15/14 (x - 3)
• y - 2 = -0.93 (x - 3)

• y - 8 = 0.2321 (x - 1)
• y - 8 = -0.2321 (x - 1)
• y - 8 = 4.308 (x - 1)
• y - 8 = -4.308 (x - 1)

Tags

CCSS.8.EE.C.7B