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What is the gradient of the tangent to the curve y = 4x^3-5x^2+x-7 at the point (2,7)?

1 Answer

Jul 18, 2018

Gradient of tangent to the curve at $(2, 7)$ $(2,7)$ is $29$ $29$

Explanation:

$y = 4 x^{3} - 5 x^{2} + x - 7; (2, 7)$ $y= 4 x^3-5 x^2+x-7 ; (2,7)$

$\frac{d y}{d x} = 12 x^{2} - 10 x + 1;$ $dy/dx = 12 x^2-10 x+1 ;$

At $x = 2, \frac{d y}{d x} = 12 \cdot 2^{2} - 10 \cdot 2 + 1 = 48 - 20 + 1 = 29$ $x=2 , dy/dx = 12* 2^2-10*2+1 =48-20+1=29$

Gradient of tangent to the curve at $(2, 7)$ $(2,7)$ is $29$ $29$ [Ans]

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