hi people, my question is: find the gradient of the curve $\displaystyle y = x^2 - 2x - 24$

a) at the point where the curve crosses the y axis
and
b) at each of the points wherethe curve crosses the x axis

for a) i differentiated (i think thats what you call what i did) like this:
$\displaystyle \frac{dy}{dx} = 2x - 2$ then made x = 0 as would be the case with the y axis and came up with - 2 which i know is right

but for b) i made y = 0 and used the quadratic formula for x.

$\displaystyle \frac{2\pm\sqrt{4 + 96}}{2a}$ and came up with x = 6 or x = -4

the answer for b) written in my textbook is -10, 10

can someone show me how i'v gone wrong please?

thankyou

2. ah sorry i've just realised how i need to finish that and its pretty simple haha. sorry for wasting your time.

3. Originally Posted by mark
ah sorry i've just realised how i need to finish that and its pretty simple haha. sorry for wasting your time.
I assume you substituted x = 6 and x = -4 into dy/dx ....

4. yep, thats what i did