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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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    but still can't solve! can you explain more? thx2012-04-10

2 Answers 2

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

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

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

$-5=4a-\frac{b}{4}\tag{3}$

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

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    @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.2012-04-10
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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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    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 $b=4$.2012-04-10