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The gradient of the tangent to the curve $y=8+4x-2x^{2}$ at the point where the curve cuts the $y$-axis is

\[\begin{array}{1 1}(1)8&(2)4\\(3)0&(4)-4\end{array}\]

1 Answer

$y= 8+4x-2x^2$
To obtain the point of intersection with $y-axis$ put $x= 0$
$y=8$
The point of intersection is (0,8)
$y= 8+4x-2x^2$
$\large\frac{dy}{dx} $$=4-4x$
$\large\frac{dy}{dx} $$/(0,8) =4-0=4$
Hence 2 is the correct answer.
answered May 19, 2014 by meena.p
 
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