Question
The curve $y=a x^{3}+b x^{2}+c x+5$ touches $x$ -axis at $A(-2,0)$. The curve intersects the $y$ -axis at a point $B$ where its slope equals 3 . The value of $^{\prime} \mathrm{a}$ ' is(a) $-2$(b) 2(c) $\frac{-1}{2}$(d) $\frac{1}{2}$
Step 1
This means that when $x=-2$, $y=0$. Substituting these values into the equation of the curve, we get: \[0 = -8a + 4b - 2c + 5 \tag{1}\] Show more…
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Implicit Differentiation
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