Solve y=x²+4x+5: Finding Values Where Function is Positive

The problem asks us to determine when the quadratic function $f(x) = x^2 + 4x + 5$ is greater than zero. Here's how we solve it:

Step 1: Analyze the Vertex
The quadratic function $f(x) = x^2 + 4x + 5$ is in the standard form $y = ax^2 + bx + c$, where $a = 1$, $b = 4$, and $c = 5$. Since $a > 0$, the parabola opens upwards, and thus the vertex represents its minimum point.
To find the x-coordinate of the vertex, use the formula $x = -\frac{b}{2a}$:

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

Substitute $x = -2$ back into the function to find the y-coordinate:

$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$

The vertex of the parabola is $(-2, 1)$, which implies the minimum value of the function is 1.

Step 2: Analyze the Discriminant
The discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots:

$\Delta = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$

Since $\Delta < 0$, the quadratic equation has no real roots, meaning it doesn't intersect the x-axis. Therefore, $f(x) > 0$ for all $x$.

Conclusion
Because the vertex is the minimum point and the function does not intersect the x-axis, the function $f(x) = x^2 + 4x + 5$ is positive for all values of $x$.

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