hi, my question is: the equation of a curve is $y = (x - 8)(x - 2)$ find the gradient of the curve
a) at the point where the curve crosses the y-axis,
b) at each of the points where the curve crosses the x-axis

for a) i expanded the brackets then differentiated:

$x^2 - 2x - 8x + 16 \implies x^2 - 10x + 16 \implies \frac{dy}{dx} = 2x - 10$ since x is zero on the y-axis the answer must be -10 and my book confirms this.

for b) i just made it $2x - 10 = 0 \implies 2x = 10 \implies x = 5$ but i don't think this is going in the right direction as the answer the book gives is "6 and -6"

could someone show me how its done please?

thanks, mark

2. Originally Posted by mark
hi, my question is: the equation of a curve is $y = (x - 8)(x - 2)$ find the gradient of the curve
a) at the point where the curve crosses the y-axis,
b) at each of the points where the curve crosses the x-axis

for a) i expanded the brackets then differentiated:

$x^2 - 2x - 8x + 16 \implies x^2 - 10x + 16 \implies \frac{dy}{dx} = 2x - 10$ since x is zero on the y-axis the answer must be -10 and my book confirms this.

for b) i just made it $2x - 10 = 0 \implies 2x = 10 \implies x = 5$ but i don't think this is going in the right direction as the answer the book gives is "6 and -6"

could someone show me how its done please?

thanks, mark
HI

y=(x-8)(x-2)

The curve cuts the x-axis at x=8 and x=2

So substitute 8 and 2 respectively into the equation of gradient u got .