Solve y = x² + 8x + 20: Finding Values Where Function is Positive

The function given is $y = x^2 + 8x + 20$. This is a quadratic function where the coefficient of $x^2$ (which is $a = 1$) is positive, indicating the parabola opens upwards.

Let’s calculate the vertex to find the minimum value of $y$. The vertex of a parabola described by $y = ax^2 + bx + c$ is found at $x = -\frac{b}{2a}$.

Here, $a = 1$, $b = 8$. So the vertex is at:

$x = -\frac{8}{2 \times 1} = -4$

Substitute $x = -4$ into the function $y = x^2 + 8x + 20$ to calculate the minimum value of $y$.

$y = (-4)^2 + 8(-4) + 20 = 16 - 32 + 20 = 4$

The minimum value of the function is $y = 4$ at $x = -4$.

Given the opening direction of the parabola and the positive minimum value, the function $f(x) = x^2 + 8x + 20$ is always greater than 0.

Thus, the function is positive for all values of $x$.