Questions: What is (are) the x-value(s) at the point(s) where y=x^2+5x+6 crosses the x-axis? (Separate multiple answers with a comma.)

Solution Steps

The given quadratic function is

\[ y = x^2 + 5x + 6 \]

To find the points where this function crosses the \(x\)-axis, we need to set \(y = 0\):

\[ x^2 + 5x + 6 = 0 \]

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by

\[ D = b^2 - 4ac \]

For our equation, \(a = 1\), \(b = 5\), and \(c = 6\):

\[ D = 5^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 \]

Since the discriminant is non-negative, we can find the roots using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

Substituting the values of \(a\), \(b\), and \(D\):

\[ x = \frac{-5 \pm \sqrt{1}}{2 \cdot 1} = \frac{-5 \pm 1}{2} \]

Calculating the two possible values for \(x\):

\[ x_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \]

\[ x_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \]

Final Answer