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the curve ax squared - 4x+1 has a stationary point where x equals -3
find the value of a and determine the nature of the stationary point

Sagot :

konrad509
[tex]f(x)=ax^2-4x+1\\ f'(x)=2ax-4\\ 0=2a\cdot(-3)-4\\ 6a=-4\\ a=-\dfrac{4}{6}=-\dfrac{2}{3}[/tex]

It's the quadratic equation where [tex]a<0[/tex] so the stationary point is a maximum.
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