The locally linear approximation of the differentiable function f at x=2 is used to approximate the value of f(2.3). The approxi
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Answer:
The answer is parabola; (y')² - 8x' = 0 ⇒ answer (b)
Step-by-step explanation:
* At first lets talk about the general form of the conic equation
- Ax² + Bxy + Cy² + Dx + Ey + F = 0
∵ B² - 4AC < 0 , if a conic exists, it will be either a circle or an ellipse.
∵ B² - 4AC = 0 , if a conic exists, it will be a parabola.
∵ B² - 4AC > 0 , if a conic exists, it will be a hyperbola.
* Now we will study our equation:
* x² - 8y = 0
∵ A = 1 , B = 0 , C =
∴ B² - 4AC = (0)² - 4(1)(0) = 0
∵ B² - 4AC = 0
∴ it will be a parabola.
∵ Ф = 90°
* The point (x , Y) will be (x' , y')
∵ x = x'cosФ - y'sinФ and y = x'sinФ + y'cosФ
∵ cos(90) = 0 and sin(90) = 1
∴ x = -y' and y = x'
* lets substitute x and y in the first equation
∴ (-y')² - 8(x') = 0
∴ (y')² - 8x' = 0
* We notice that the x' took the place of y and y' took the place of x
∴ The parabola rotated around the origin by 90°
∴ The equation of the parabola is (y')² - 8x' = 0
* The answer is parabola, with angle of rotation 90°
* The equation is (y')² - 8x' = 0
* Look to the graph
- The blue is x² - 8y = 0
- The green is (y')² - 8x' = 0
