Problem

Find the range of the function \( f(x) = x^2 + 5x + 6 \).

Answer

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Answer

Step 3: Because the coefficient of \( x^2 \) is positive, the parabola opens upwards. This means that the vertex is the minimum point of the function, and the range of the function is \( y \geq -0.25 \).

Steps

Step 1 :Step 1: To find the range of the function, we first need to find the vertex of the parabola \( f(x) = x^2 + 5x + 6 \). The x-coordinate of the vertex can be found using the formula \( -b/2a \), where a and b are the coefficients of \( x^2 \) and \( x \) respectively. In this case, a=1 and b=5, so the x-coordinate of the vertex is \( -5/2(-1) \), which simplifies to \( -5/2 \).

Step 2 :Step 2: To find the y-coordinate of the vertex, we substitute \( x = -5/2 \) into the equation. This gives us \( f(-5/2) = (-5/2)^2 + 5(-5/2) + 6 \), which simplifies to \( 6.25 - 12.5 + 6 \), or \( -0.25 \).

Step 3 :Step 3: Because the coefficient of \( x^2 \) is positive, the parabola opens upwards. This means that the vertex is the minimum point of the function, and the range of the function is \( y \geq -0.25 \).

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