Find the equation of the tangent to the curve y = 3x^2(x+2)^6 at the point (-1,3), in the form y = mx+c
Curve: y = 3x2(x+2)6 Coordinate: (-1, 3)
This is typically a C3/4 level question because of the differentiation, but the rest of the question is possible with year 12 maths knowledge. The best way to tackle this is to find the gradient function of the curve by differentiating, this will give us the gradient of the curve at (-1,3), which is equal to the gradient of the tangent at (-1,3). We then use the equation y-y1=m(x-x1) (where (x1,y1) = (-1,3) ) to find the equation of the tangent.
Differentiate using the product rule. dy/dx = vu' + uv'
u = 3x2, v = (x+2)6, u' = 6x, v' = 6(x+2)5
dy/dx = ( (x+2)6 *6x ) + ( 3x2 * 6(x+2)5) = 6x(x+2)6 + 18x2(x+2)5
When x = -1,
dy/dx = ((6 * -1)(-1 + 2)6) + ((18 * 1) * (-1 + 2)5) = -6 + 18 = 12
y - 3 = 12(x+1)
y = 12x+12+3
y = 12x + 15
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