# Find 2 coordinates of the points where the function has a horizontal tangent line? 2y^2+4x^2-y=16x

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Feb 9, 2018

Tangent is horizontal at $\left(2 , 3.09\right)$ and $\left(2 , - 2.59\right)$

#### Explanation:

The function $2 {y}^{2} + 4 {x}^{2} - y = 16 x$ will have a horizontal tangent line, where slope of the curve i.e. derivative of function equals zero.

Hence let us first find the derivative of $2 {y}^{2} + 4 {x}^{2} - y = 16 x$. Differentiating (considering it an implicit function), we get

$4 y \frac{\mathrm{dy}}{\mathrm{dx}} + 8 x - \frac{\mathrm{dy}}{\mathrm{dx}} = 16$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left(4 y - 1\right) = 16 - 8 x$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{16 - 8 x}{4 y - 1} = \frac{8 \left(2 - x\right)}{4 y - 1}$

and this is zero when $x = 2$.

When $x = 2$, we have $2 {y}^{2} + 4 \cdot {2}^{2} - y = 16 \times 2$

or $2 {y}^{2} - y + 16 - 32 = 0$

or $2 {y}^{2} - y - 16 = 0$

i.e. $y = \frac{1 \pm \sqrt{1 + 128}}{4} = \frac{1 \pm \sqrt{129}}{4}$

i.e. $3.09$ or $- 2.59$

Hence tangent is horizontal at $\left(2 , 3.09\right)$ and $\left(2 , - 2.59\right)$

graph{(2y^2+4x^2-y-16x)((x-2)^2+(y-3.09)^2-0.02)((x-2)^2+(y+2.59)^2-0.02)=0 [-10, 10, -5, 5]}

• 30 minutes ago
• 30 minutes ago
• 31 minutes ago
• 31 minutes ago
• 7 minutes ago
• 9 minutes ago
• 10 minutes ago
• 14 minutes ago
• 25 minutes ago
• 30 minutes ago
• 30 minutes ago
• 30 minutes ago
• 31 minutes ago
• 31 minutes ago