Question
$y=x^2$, translated so that the minimum occurs at $(-4,-8)$
Step 1
The function \( y = x^2 \) has its vertex at the point \( (0, 0) \). Show more…
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If $4 x^{2}+8 x y+9 y^{2}-8 x-24 y+4=0$, show that when $\frac{\mathrm{d} y}{\mathrm{~d} x}=0, x+y=1$ and $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{4}{8-5 y}$. Hence find the maximum and minimum values of $y$.
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Further problems F.8
If $4 x^{2}+8 x y+9 y^{2}-8 x-24 y+4=0$, show that when $\frac{d y}{d x}=0, x+y=1$ and $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=\frac{4}{8-5 y}$. Hence find the maximum and minimum values of $y$.
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Determine the maximum or minimum $y$ -value. $$ y=-x 2-4 x+8 $$
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