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The equation of tangent to the curve y=x...

The equation of tangent to the curve `y=x^(2)+4x+1` at (-1, -2) is

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Given , equation of curve is ` y = x^(2) + 4x + 1 `
` therefore (dy)/(dx) = 2x + 4 = 2 (x + 2)`
i.e., ` m = 2 (x + 2)`
` therefore m at (-1, -2) ` will be
` m = 2 (-1 + 2) = 2 xx 1`
` therefore m = 2 `
Now , equation of tangent at point(x, y) with slope m is
` y - y_(1) = m (x - x_(1))`
Now , at point `(-1 , -2) ` , equations of tangent will be ,
` rArr y + 2 = 2 (x + 1)`
` rArr y + 2 = 2x + 2 `
` rArr y = 2x or y- 2x = 0 ` will be the
equation of tangent to the given curve .
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