Determine the x-intercept of the quadratic function:
y=-0.25(x+1)^2+9

Question

14-09-2015
Mathematics

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Determine the x-intercept of the quadratic function:
y=-0.25(x+1)^2+9

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Panoyin Panoyin · Answer 1 · 2015-09-14T02:16:31+02:00

y = ⁻¹/₄(x + 1)² + 9
0 = ⁻¹/₄(x + 1)² + 9
0 = ⁻¹/₄(x + 1)(x + 1) + 9
0 = ⁻¹/₄(x(x + 1) + 1(x + 1)) + 9
0 = ⁻¹/₄(x(x) + x(1) + 1(x) + 1(1)) + 9
0 = ⁻¹/₄(x² + x + x + 1) + 9
0 = ⁻¹/₄(x² + 2x + 1) + 9
0 = ⁻¹/₄(x²) - ¹/₄(2x) - ¹/₄(1) + 9
0 = ⁻¹/₄x² - ¹/₂x - ¹/₄ + 9
0 = ⁻¹/₄x² - ¹/₂x + 8³/₄
x = -(⁻¹/₂) ± √((⁻¹/₂)² - 4(⁻¹/₄)(8³/₄))
   2(⁻¹/₄)
x = ¹/₂ ± √(¹/₄ + 8³/₄)
   ⁻¹/₂
x = ¹/₂ ± √(9)
      ⁻¹/₂
x = ¹/₂ ± 3
      ⁻¹/₂
x = ¹/₂ + 3 U x = ¹/₂ - 3
      ⁻¹/₂    ⁻¹/₂
x = 4 U x = -2¹/₂
    ⁻¹/₂ ⁻¹/₂
x = -8 U x = 5

Determine the x-intercept of the quadratic function: y=-0.25(x+1)^2+9

Respuesta :

Otras preguntas

Determine the x-intercept of the quadratic function:
y=-0.25(x+1)^2+9