for the curve with equation y=(x^2+1)/(x^2-4), find (i) the cordinates of the turning point(s) (ii) the equations of the asymptotes (iii) sketch the curve
To find the coordinates of the turning point(s) of the curve with equation y = (x^2 + 1)/(x^2 - 4), we need to determine the critical points of the function.
(i) Finding the Turning Point:
1. Start by finding the derivative of the given function. We can use the quotient rule to differentiate it.
Let's denote the derivative as dy/dx.
y = (x^2 + 1)/(x^2 - 4)
dy/dx = [(x^2 - 4)(2x) - (x^2 + 1)(2x)] / (x^2 - 4)^2
Simplify dy/dx to get:
dy/dx = [(2x^3 - 8x - 2x^3 - 2x)] / (x^2 - 4)^2
dy/dx = [-10x] / (x^2 - 4)^2
2. The turning points occur where the derivative is equal to zero. Set dy/dx = 0 and solve for x:
-10x = 0
x = 0
3. Now that we have x = 0, substitute it back into the original equation to find the corresponding y-value:
y = (0^2 + 1) / (0^2 - 4)
y = 1 / -4
y = -1/4
Therefore, the coordinates of the turning point are (0, -1/4).
(ii) Finding the Equations of the Asymptotes:
1. First, determine the vertical asymptotes by identifying any values of x that make the denominator of the original equation equal to zero.
x^2 - 4 = 0
(x - 2)(x + 2) = 0
x = 2 or x = -2
Hence, the vertical asymptotes of the curve occur at x = 2 and x = -2.
2. For the horizontal or slant asymptote, we examine the behavior of the curve as x approaches positive or negative infinity.
As x approaches infinity:
y = (x^2 + 1) / (x^2 - 4) ≈ x^2 / x^2 = 1
As x approaches negative infinity:
y = (x^2 + 1) / (x^2 - 4) ≈ x^2 / x^2 = 1
Therefore, the curve has a horizontal asymptote at y = 1.
(iii) Sketching the Curve:
Plot the coordinates of the turning point (0, -1/4) on a graph. Draw the vertical asymptotes at x = 2 and x = -2. Finally, sketch a smooth curve that approaches the horizontal asymptote y = 1 as x increases or decreases, passing through the turning point.