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The derivative gives the slope at the point.
y' = 2 - 3x^2
y'(1) = 2 - 3 = -1
The tangent line can be represented by the equation
y = -x + constant (slope-intercept form)
x + y = constant (standard form)
x + y = 3 (the constant is 1+2 = 3)
y' = 2 - 3x^2
y'(1) = 2 - 3 = -1
The tangent line can be represented by the equation
y = -x + constant (slope-intercept form)
x + y = constant (standard form)
x + y = 3 (the constant is 1+2 = 3)
0
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