Calculas1?
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested.
y 6 + x 3 = y 2 + 9x, slope at (0, 1)
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve, as requested.
y 6 + x 3 = y 2 + 9x, slope at (0, 1)
Captain Matticus, LandPiratesInc
y^6 + x^3 = y^2 + 9x
6y^5 * dy + 3x^2 * dx = 2y * dy + 9 * dx
x = 0 , y = 1
6 * dy + 0 * dx = 2 * dy + 9 * dx
4 * dy = 9 * dx
dy/dx = 9/4
Tangent slope: 9/4
Normal slope: -4/9
y - 1 = m * (x - 0)
y - 1 = mx
y = mx + 1
Tangent line: y = (9/4) * x + 1
Normal line: y = (-4/9) * x + 1