How do you find the slope of the tangent line $y=x^3-7x^2+4x+9$ at x=1?

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Answer 1 · 2016-12-26T19:20:29

$-7$

The slope of the tangent line at a particular point is the gradient of the curve at that point. So we differentiate

$y=x^3-7x^2+4x+9$

$(dy)/(dx)=3x^2-14x+4$

$((dy)/(dx))_(x=1)=3xx1^2-14xx1+4$

$((dy)/(dx))_(x=1)=3-14+4=-7$

How do you find the slope of the tangent line y=x^3-7x^2+4x+9y=x^3-7x^2+4x+9 at x=1?