To determine the maximum value of the function $y = x^2 + 10x + 1$, we need to follow these steps:

Step 1: Identify the type of the function

The given function $y = x^2 + 10x + 1$ is a quadratic function of the form $y = ax^2 + bx + c$. In this case:

• $a = 1$ (positive, indicating that the parabola opens upwards),
• $b = 10$,
• $c = 1$.

Since $a = 1 > 0$, the function has a minimum value, not a maximum. So, instead of finding the maximum, we will determine the minimum value of this function.

Step 2: Find the vertex of the parabola

The vertex of a parabola given by $y = ax^2 + bx + c$ occurs at the x-coordinate: $x = -\frac{b}{2a}$ Substituting $b = 10$ and $a = 1$ into the formula: $x = -\frac{10}{2(1)} = -5$

Step 3: Calculate the minimum value of $y$

Now, substitute $x = -5$ into the function $y = x^2 + 10x + 1$: $y = (-5)^2 + 10(-5) + 1 = 25 - 50 + 1 = -24$

Thus, the minimum value of the function is $y = -24$, and it occurs when $x = -5$.

Conclusion:

The function does not have a maximum value, but its minimum value is $y = -24$, at $x = -5$.

Would you like further details on this or have any questions?

Related questions:

1. How would the minimum value change if the constant term were different?
2. What if $a$ in the quadratic function was negative—how would the function behave?
3. Can you find the axis of symmetry for this function?
4. How do we find the roots of this quadratic function?
5. How does completing the square help in finding the vertex?

Tip:

For any quadratic function $y = ax^2 + bx + c$, the vertex formula $x = -\frac{b}{2a}$ is a quick way to find where the maximum or minimum occurs.