How to Sketch a Parabola - Example 3 (y = x^2 + 6x + 10)

Graphing the function

y = x² + 6x + 10

we go through the normal process of finding:

• its concavity (whether it smiles or frowns)
• y-intercept
• turning point / vertex
• any x-intercepts

But this particular example does not have any x-intercepts because as a positive parabola, the y-coordinates for the vertex and intercepts are positive, and therefore, the parabola will not cross the x-axis.

And also, by the quadratic formula, which is used to solve for the x-intercepts:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

The term b² - 4ac evaluates to a negative number, which cannot be calculated in terms of real numbers.