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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.

I did this way:
$$-2=a(2)^2+\frac{b}{2}$$ $$-2=4a+\frac{b}{2}$$ x 2 $$-4=8a+b$$ $$-\frac{1}{2}a+\frac{b}{8}$$ But the answer is $a=-1$, $b=4$.

Could you help me out?
(this chapter is about differentiation of polynomials, power functions and rational functions)

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Algebra error: when you multiply $-2=4a+\frac{b}2$ by $2$ you should get $-4=8a+b$, and if you then divide by $8$, you should get $-\frac12=a+\frac{b}8$. (I would not divide by $8$.) – Brian M. Scott Apr 10 '12 at 5:40
You need also to deal with the gradient part. We have $\frac{dy}{dx}=2ax-\frac{b}{x^2}$. When $x=2$, this is $-5$. – André Nicolas Apr 10 '12 at 5:44
but still can't solve! can you explain more? thx – Sb Sangpi Apr 10 '12 at 6:05
up vote 5 down vote accepted

$$y=ax^2+\frac{b}{x}$$ $$\stackrel{\text{(2,-2)}}{\implies}-2=a(2^2)+\frac{b}{2}\tag{1}$$

We also have: (derivative) $y'=2ax-\frac{b}{x^2}\tag{2}$

It has a gradient of $-5$ at that point which means: $$(2)\stackrel{\text{(2,-5)}}{\implies}-5=2a(2)-\frac{b}{2^2}$$


$(1)$ and $(3)$ $\implies$ $a=-1$ and $b=4$

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$-2=8a+b$ and $-5=16a-b$ but still can't get $a=-1$ and $b=4$ sorry about that! – Sb Sangpi Apr 10 '12 at 12:36
@SbSangpi: You've got wrong equations. $-2=a(2^2)+\frac{b}{2}$ would be (multiplying both sides by 2) $-4=8a+b$. Do you see how? – Gigili Apr 10 '12 at 13:31
@SbSangpi: The other equation $-5=4a-\frac{b}{4}$ would be $-20=16a-b$ multiplying by 4. Now you should add the two equation: $-20 \color{red}{-4}=16a-b +\color{red}{8a+b}$. Let me know if there are any other problems. – Gigili Apr 10 '12 at 13:58

We must have a system of equations to solve for $a$ and $b$. In the above comment, we have the equation $8a+b = -4$ and $16a-b= -20$. By elimination, we have $a=-1$ and $b=4$.

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how did $8a+b=-4$ into $16a-b=-20$? – Sb Sangpi Apr 10 '12 at 6:08
The $16a-b=-20$ comes from the gradient calculation, which gives $4a-\frac{b}{4}=-5$. Multiply through by $4$ to clear fractions. – André Nicolas Apr 10 '12 at 6:24
then where is $8a+b=-4 $ gone?, sorry for my lack of maths. – Sb Sangpi Apr 10 '12 at 6:31
We have two equations in the two unknowns $a$ and $b$. In simplified form they are $8a+b=-4$ and $16a-b=-20$. To solve, there are various methods. You could use the second equation to get $b=8a+20$, and substitute $8a+20$ for $b$ in the first equation. Or more simply, "add" the two equations. We get $(8a+b)+(16a-b)=-4+(-20)$, so $24a=-24$, $a=-1$. Then use $8a+b=-4$, with $a=-1$, to get $b=4$. – André Nicolas Apr 10 '12 at 6:38

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