Hi
I few question i am having troubling finding the gradient:

1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

this is how i approached it:
$\frac{dy}{dx}=2ax+bx^-2$
when x=2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-5$(1)

when x=2,y=-2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-2$(2)

(1)-(2)
this is where a got stuck because when i added and minus it didn't work.

2)Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

P.S

2. Originally Posted by Paymemoney
Hi
I few question i am having troubling finding the gradient:

1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

this is how i approached it:
$\frac{dy}{dx}=2ax+bx^-2$
when x=2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-5$(1)

when x=2,y=-2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-2$(2)

(1)-(2)
this is where a got stuck because when i added and minus it didn't work.

2)Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

P.S
1) You know that $(2, -2)$ lies on the curve $y = ax^2 + \frac{b}{x}$.

So $-2 = a(2)^2 + \frac{b}{2}$

$-2 = 4a + \frac{b}{2}$

$-4 = 8a + b$

$b = -8a - 4$.

You also know that the gradient is $-5$ at this point.

So $\frac{dy}{dx} = 2ax - \frac{b}{x^2}$

$-5 = 2a(2) - \frac{b}{2^2}$

$-5 = 4a - \frac{b}{4}$

$-20 = 16a - b$

$b = 16a - 20$.

But since $b = -8a - 4$ as well, we have

$-8a - 4 = 16a - 20$

$16 = 24a$

$a = \frac{2}{3}$.

And since $b = -8a - 4$

$b = -8\left(\frac{2}{3}\right) - 4$

$b = -\frac{16}{3} - 4$

$b = -\frac{16}{3} - \frac{12}{3}$

$b = -\frac{28}{3}$.

3. Originally Posted by Paymemoney
Hi
I few question i am having troubling finding the gradient:

1) Given that the curve y=ax^2+\frac{b}{x} has a gradient of -5 at the point (2, -2) find the value of a and b.

this is how i approached it:
$\frac{dy}{dx}=2ax+bx^-2$
when x=2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-5$(1)

when x=2,y=-2
$\frac{dy}{dx}= 4a-\frac{b}{4}=-2$(2)

(1)-(2)
this is where a got stuck because when i added and minus it didn't work.

2)Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the x-axis.

P.S
2) the curve crosses the x-axis at the point where $y=0$. find $dy/dx$ at those point.