To find the range of the quadratic function \( y = 8x^2 + 32x + 24 \), we first need to identify its vertex, since the vertex will provide information about the minimum or maximum point of the parabola.
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Identify the coefficients: The given quadratic function is in the standard form \( y = ax^2 + bx + c \), where:
- \( a = 8 \)
- \( b = 32 \)
- \( c = 24 \)
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Locate the vertex: The x-coordinate of the vertex of a quadratic function can be found using the formula: \[ x = -\frac{b}{2a} \] Plugging in the values for \( b \) and \( a \): \[ x = -\frac{32}{2 \cdot 8} = -\frac{32}{16} = -2 \]
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Find the corresponding y-coordinate: Now, we substitute \( x = -2 \) back into the quadratic function to find the minimum value of \( y \): \[ y = 8(-2)^2 + 32(-2) + 24 \] \[ = 8 \cdot 4 - 64 + 24 \] \[ = 32 - 64 + 24 \] \[ = -8 \]
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Determine the direction of the parabola: Since the coefficient \( a \) (which is 8) is positive, the parabola opens upwards. This means that the vertex at \( (-2, -8) \) is the minimum point of the function.
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Find the range: The range of the quadratic function is all values of \( y \) greater than or equal to the minimum value at the vertex. Thus, the range is given by: \[ y \geq -8 \] In interval notation, the range is: \[ [-8, \infty) \]
Therefore, the range of the quadratic function \( y = 8x^2 + 32x + 24 \) is \(\boxed{[-8, \infty)}\).