WAEC - Further Mathematics (2010 - No. 19)
Find the equation of the tangent to the curve \(y = 4x^{2} - 12x + 7\) at point (2, -1).
y + 4x - 9 = 0
y - 4x - 9 = 0
y - 4x + 9 = 0
y + 4x + 9 = 0
คำอธิบาย
\(y = 4x^{2} - 12x + 7\)
\(\frac{\mathrm d y}{\mathrm d x} = 8x - 12\)
At x = 2, y = 8(2) - 12 = 4
Equation of the tangent to the curve: \(y - (-1) = 4(x - 2)\)
\(y + 1 = 4x - 8 \implies y - 4x + 1 + 8 = y - 4x + 9 = 0\)