This seems simple but it doesnt really sink in for me.
Can i have working out plz. thanx for any help
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.