This seems simple but it doesnt really sink in for me.

http://img205.imageshack.us/img205/8137/1241413ii3.jpg

Can i have working out plz. thanx for any help

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- Oct 25th 2007, 04:02 AMKavXfind gradient, equation, value, graph included.
This seems simple but it doesnt really sink in for me.

http://img205.imageshack.us/img205/8137/1241413ii3.jpg

Can i have working out plz. thanx for any help - Oct 25th 2007, 08:57 AMhoranSolution
I'm not sure about (i), but you don't need it to find the other two answers.

ii. The slope of the tangent at a is given by dy/dx at a. So $\displaystyle y'=2x$, or, at a, $\displaystyle y'=2a$. We also know that the value of y(a) is a^2. Using point-slope form, the equation of the tangent is given by:

$\displaystyle (y-y_0)=m(x-x_0)$

or $\displaystyle (y-a^2)=2a(x-a)$

iii. Plugging in the point (2,0) into the equation from (ii), we get

$\displaystyle (0-a^2)=2a(2-a)$

Then:

$\displaystyle -a^2=4a-2a^2$

$\displaystyle 0=4a-a^2$

$\displaystyle 0=-a(a-4)$

$\displaystyle a=[0,4]$

Since a > 2 by the graph, a must be equal to 4.