MATH SOLVE

5 months ago

Q:
# For A, is it correct that the roots/x-intercepts are the positive and negative square root of 7?For B, how could I prove that the tangents are parallel? The wording reminds me of the Mean Value Theorem, but we didn't spend a lot of time on that in class and I'm not terribly familiar with it.

Accepted Solution

A:

A. The x-intercept is where y=0.

x^2 + 0 + 0 = 7

So yes, x = +/-sqrt(7)

B. The equation is an eclipse.

+/- sqrt(7) = +/- 2.64....

mean value theorem will tell you its continuous and differentiable, Use points on either side of +/- 2.64

(2,1) (sqrt7, 0) (3,-1)

slope: (-1-1)/(3-2) = -2

(-2,-1) (-sqrt7,0) (-3,1)

slope: (1+1)/(-3+2) = -2

If this is for calculus, take derivative at (0,sqrt7)

2x + y + x(dy/dx) + 2y(dy/dx) = 0

2(sqrt7) + 0 + (sqrt7)(dy/dx) + 0 = 0

-2 = dy/dx

likewise for x = -sqrt7

dy/dx = -2

or use the limit definition of derivative with h, as h --> zero. give tangent at specific point.

A. The x-intercept is where y=0.

x^2 + 0 + 0 = 7

So yes, x = +/-sqrt(7)

B. The equation is an eclipse.

+/- sqrt(7) = +/- 2.64....

mean value theorem will tell you its continuous and differentiable, Use points on either side of +/- 2.64

(2,1) (sqrt7, 0) (3,-1)

slope: (-1-1)/(3-2) = -2

(-2,-1) (-sqrt7,0) (-3,1)

slope: (1+1)/(-3+2) = -2

If this is for calculus, take derivative at (0,sqrt7)

2x + y + x(dy/dx) + 2y(dy/dx) = 0

2(sqrt7) + 0 + (sqrt7)(dy/dx) + 0 = 0

-2 = dy/dx

likewise for x = -sqrt7

dy/dx = -2

or use the limit definition of derivative with h, as h --> zero. give tangent at specific point.