If the points (X, Y) given in the following table are joined serially on X – Y plane then we getX –2 –1 0 1 2 Y 4 1 0 1 4
Parabola
To identify the curve, we look for a rule that connects X and Y for every point in the table. If each Y value equals the square of the corresponding X value, the points satisfy \(Y = X^{2}\), which is a second-degree equation whose graph is a parabola symmetric about the Y-axis.
Check each point against \(Y = X^{2}\):
\(X = -2: Y = (-2)^{2} = 4\)
\(X = -1: Y = (-1)^{2} = 1\)
\(X = 0: Y = 0^{2} = 0\)
\(X = 1: Y = 1^{2} = 1\)
\(X = 2: Y = 2^{2} = 4\)
All five points satisfy \(Y = X^{2}\), with vertex (0, 0) and the curve opening upward.
A straight line would need a constant change in Y for equal changes in X, but here Y falls and then rises. A hyperbola such as XY = constant cannot pass through the origin, and a circle would give two Y values for most X values and could not keep rising on both sides. The symmetric U shape fits only a parabola.
Hence, Parabola.
Important Questions from Quadratic Equation
- If \(\rm \left( \frac{x}{x+1} \right)^2 -5 \left( \frac{x}{x+1} \right) +6=0 \) , then the value of \(\rm \left( 1+\frac{1}{x} \right) \) is equal to :
If the roots of the equation (q – r)x 2+ (r – p)x + (p – q) = 0 are equal, then which of the following is true?
Number of real roots of the quadratic equation 3x 2+ 4x + 25 = 0 is
The value of m for which one of the root of x 2 - 3x + 2m = 0 is double of the root of x 2 - x + m = 0 is:
If x 2+ 1 = 2x, then find x – \((\frac{1}{x})\)
