Find the point(s) if any of horizontal tangent lines for the function: x^2 + xy + y^2 = 6?

Answer 1

horizontal line has zero slope.

f(x) = x^2 + xy + y^2 = 6

f'(x) = slope of the tangent

f'(x) = 2x + y + x y' + 2 yy' = 0

y'(x + 2y) = - (2x+y)

y' = -(2x + y)/(x+2y)

since slope = 0

-(2x + y)/(x+2y) = 0

-2x - y = 0

y = - 2x

substituting this in the function

x^2 + x(-2x) + (-2x)^2 = 6

x^2 - 2x^2 + 4x^2 = 6

3x^2 = 6

x^2 = 6/3 = 2

x = +/- sqrt(2)

y = +/- 2 sqrt(2)

The point of tangency is [+/- sqrt(2), +/- 2 sqrt(2)]

the equation of tangent is

y - y1 = m(x - x1)

y +/- 2 sqrt(2) = 0

y = +/- 2 sqrt(2) are the horizontal tangent lines to the

given function

Answer 2

The function, x^2 ? xy ? y^2 = 5, is a hyperbola, meaning there would desire to be 2 factors the place it curves around and subsequently has 2 vertical tangent lines. (while you're allowed to graph, as in AP Calc, you will see this.) subsequently, basically B or E is right. we are going to examine the values for B. to locate the vertical tangent lines, you will locate the place the decrease of the spinoff techniques infinity... Differentiate implicitly, (surely multiply the two sides by way of d/dx) x^2 ? xy ? y^2 = 5 and you get 2x - y - (dy/dx)x - 2(dy/dx)y = 0, or dy/dx = (2x - y) / (x + 2y). This techniques infinity if (2x - y) is non-0 and (x + 2y) = 0. as a result, x = -2y on the tangent line. considering that -5y isn't unavoidably 0, we are going to proceed. pondering selection B, (2, -a million) looks to artwork, considering that -2y = x and -2(-a million) = 2. in spite of the undeniable fact that, -2(-2) =/= a million. the different actually answer is (-2, a million). subsequently, the respond is E.

Answer 3

implicit derivative
2x+y+xy´+2yy´= 0
y´= (-2x-y)/(x+2y)=0 so
2x+y=0
y=-2x and
x^2-2x^2 +4x^2=6
3x^2= 6 ans x=+-sqrr(2)and y = -+2sqrt(2)
(sqrt(2),-2sqrt2) and (-sqrt(2),+2sqrt(2))