Question

calculus question!
find the points at which the graph of the equation has a vertical and or horizontal tangent line.
5x^2+8y^2+10x+48y+8=0

Answer 1

Use implicit differentiation to get:
\[10x+16y \frac{ dy }{ dx }+10+48 \frac{ dy }{ dx }=0\]
Factor out dy/dx and rearrange to get:
\[\frac{ dy }{ dx }=\frac{ -10(x+1) }{ 16(y+3) }\]
Horizontal tangents occur when numerator of rational expression on right is zero, that is when x=-1. Substitute this value into the original equation and solve for the y values.
Similarly for vertical tangents which occur when the denominator is zero for y=-3 and substitute this into the original equation to find corresponding x values.