Respuesta :

caylus
Hello,

[tex]y=2x+b \\\\ y=ax^3\\\\ slope\ =\ 3ax^2\\ if \ x=5\ then\ slope=75a\ =2\\\\ a= \dfrac{2}{75}\ and \ y= \dfrac{2}{75}* x^3\\\\ if \ x=5\ then\ y=\dfrac{2}{75}* 5^3= \dfrac{10}{3} \\\\ \dfrac{10}{3}=2*5+b\\\\ b= \dfrac{-20}{3} [/tex]
meerkat18
the tangent of the curve is determined by getting the first derivative of teh equation of the curve and substituting with  the given data. in this case, the derivative of y=ax^3 is y' = 3a x^2. when x is equal to 5, y' = 3a (25) = 75 a

x=5; y = 125 a
y-y1 = m(x-x1)
y-125 a = 75 a (x-5)
y = 75 ax -500

-2x + y = b
-2(75/2) x  + y = -500
a = 75/2
b = -500

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