Question
$y=a x^{2}+b x+c$$\frac{d y}{d x}=2 a x+b=10 a+b=0 \quad(a t 5,4)$$\mathrm{Now}_{3} \cdot 4=25 a+5 \mathrm{~b}+\mathrm{c}$using (I) $4=25 a+5(-10 a)+c$$c=4+25 a$$\Rightarrow$ Maximum value of $c=104$
Step 1
Step 1: We are given the equation of a curve $y=a x^{2}+b x+c$ and its derivative $\frac{d y}{d x}=2 a x+b=10 a+b=0$ at the point $(5,4)$. Show more…
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