Question 1: 10 Pts. Find Ax2bxc Where A=15, B=15

Question 1 10ptsfx Ax2bxc Where A 15 B 15 An

Question 1 10 pts f ( x ) = ax 2 + bx + c , where a = -15, b = 15 and c = 14. If then h

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This problem involves analyzing a quadratic function of the form f(x) = ax² + bx + c, with specific coefficients a = -15, b = 15, and c = 14. The goal is to find the value of k such that x lies between h and k, where the function's behavior—and likely its roots or vertex—dictates these bounds. To accurately determine k, we first analyze the quadratic's properties, such as vertex, axis of symmetry, and roots.

Given the quadratic function f(x) = -15x² + 15x + 14, the first step is to find its roots where f(x) = 0. Using the quadratic formula:

x = [-b ± √(b² - 4ac)] / (2a)

we substitute a = -15, b = 15, c = 14:

x = [-15 ± √(15² - 4(-15)14)] / (2 * -15)

Calculating discriminant:

Δ = 225 - 4  -15  14 = 225 + 840 = 1065

The square root of discriminant:

√1065 ≈ 32.63

Now, compute the roots:

x₁ = [-15 + 32.63] / (-30) ≈ 17.63 / -30 ≈ -0.5877


x₂ = [-15 - 32.63] / (-30) ≈ -47.63 / -30 ≈ 1.5877

Since the parabola opens downward (a 

x_vertex = -b/(2a) = -15 / (2 * -15) = -15 / -30 = 0.5

At x = 0.5, the parabola attains its maximum value.

Given the roots and the vertex, the parabola is positive between the roots and negative outside. Depending on the context, the interval of interest for x where f(x) is positive or negative could be between the roots or outside them. If the question's focus is to find the value of k where these properties hold, the critical value is close to the larger root, approximately 1.59.

Therefore, the value of k to two decimal places is approximately 1.59.

References

  • Anton, H., Bivens, I., & Davis, S. (2013). Calculus (10th ed.). Wiley.
  • Larson, R., & Edwards, B. H. (2014). Calculus of a Single Variable (10th ed.). Cengage Learning.
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