Find the tangent line to f(x)=3x^5-4x^2+9x-12 at x=-1
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By differentiating, we get dy/dx = 15x⁴-8x+9
The slope at x = -1 is 15+8+9=32
The value of y at x = -1 is -3-4-9-12 = -28
The equation of tangent is 32(x+1) = (y+28)
On simplifying, we get 32x - y + 4 =0
The slope at x = -1 is 15+8+9=32
The value of y at x = -1 is -3-4-9-12 = -28
The equation of tangent is 32(x+1) = (y+28)
On simplifying, we get 32x - y + 4 =0
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