<p>The graph of the quadratic trinomial \(y = ax^2 + bx + c\) has its vertex at \((4, -5)\) and two \(x\)-intercepts, one positive and one negative. Which of the following holds good?</p>
Step-by-Step Solution
Key Concept: Use vertex form y = a(x-h)² + k to establish that a > 0 (parabola opens upward since vertex is minimum with two x-intercepts). The x-intercepts being one positive and one negative means the parabola crosses both sides of the y-axis, which requires c < 0 (y-intercept is negative).
<p><strong>Step 1:</strong> Vertex at (4, -5) with two x-intercepts means the parabola has a minimum point below the x-axis and crosses it twice. This requires <strong>a > 0</strong> (parabola opens upward).</p><p><strong>Step 2:</strong> Since one x-intercept is positive and one is negative, the parabola crosses the y-axis below the origin. Thus <strong>c = f(0) < 0</strong>.</p><p><strong>Step 3:</strong> From vertex form: y = a(x-4)² - 5. The axis of symmetry is x = 4, and since roots are on opposite sides of origin, by Vieta's formulas: if roots are r₁ < 0 and r₂ > 0, then r₁·r₂ = c/a < 0, confirming c < 0 (since a > 0).</p><p><strong>Step 4:</strong> For the vertex x-coordinate: -b/(2a) = 4, so <strong>b = -8a < 0</strong> (since a > 0).</p><p><strong>Step 5:</strong> Therefore: <strong>a > 0, b < 0, c < 0</strong>. Additional valid statements typically include: b² - 4ac > 0 (two real roots), f(0) < 0, and a + b + c < 0 (since f(1) < 0).</p><p>∴ Typically: a > 0, b < 0, c < 0, and b² - 4ac > 0</p>
Correct Answer: A,B,C,D