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To determine the range of values for the variable x where the linear equation y = 5x – 6 is situated beneath the quadratic curve represented by y = x^2, one must first establish the points of intersection between the two equations. This involves setting the two expressions for y equal to each other, leading to the equation 5x – 6 = x^2. Rearranging this equation results in a standard quadratic form, specifically x^2 – 5x + 6 = 0. The next step is to factor this quadratic equation, which can be expressed as (x – 2)(x – 3) = 0, yielding the solutions x = 2 and x = 3.
Having identified the intersection points, the next phase is to analyze the intervals created by these points to ascertain where the linear function is less than the quadratic function. The intervals to consider are x < 2, 2 < x < 3, and x > 3. By selecting test values from each interval and substituting them back into the original equations, one can evaluate whether the linear equation yields a value that is less than that of the quadratic equation. For instance, choosing a value such as x = 1 for the interval x < 2 results in y = 5(1) – 6 = -1 for the linear equation and y = (1)^2 = 1 for the quadratic equation, confirming that the line is indeed below the curve in this interval.
Consequently, the analysis reveals that the linear function y = 5x – 6 remains beneath the quadratic function y = x^2 for the interval x < 2 and the interval x > 3. Therefore, the complete set of values for x where the line lies below the curve is expressed as x < 2 or x > 3. For students seeking assistance in understanding these concepts, an A Level Maths Tutor can provide valuable guidance in mastering the techniques required for such analyses.
