The given function is $y = a(x + p)^2 + q$, where $a > 0$ and $q > 0$.

This represents a parabola that opens upwards because $a > 0$. Let's analyze the number of roots (i.e., the number of $x$-intercepts, which occur where $y = 0$).

Step 1: Set $y = 0$ to find the roots

$0 = a(x + p)^2 + q$ $a(x + p)^2 = -q$ $(x + p)^2 = -\frac{q}{a}$

Step 2: Analyze the equation

• The left side, $(x + p)^2$, is always non-negative for any real number $x$ because it's a square term.
• The right side, $-\frac{q}{a}$, is negative since $q > 0$ and $a > 0$, making the expression $-\frac{q}{a} < 0$.

Conclusion:

There are no real solutions to this equation, because a square term cannot be negative. Therefore, the function has:

• No real roots.

The parabola does not intersect the $x$-axis because it is entirely above the axis due to the positive values of both $a$ and $q$.

Do you want more details, or do you have any questions?

Here are some related questions to think about:

1. What if $q = 0$? How many roots does the function have then?
2. What happens to the roots if $a < 0$ and $q > 0$?
3. Can you interpret the vertex of this parabola in terms of $p$ and $q$?
4. How does changing $p$ affect the position of the parabola?
5. How would you find the minimum value of this parabola?

Tip: The number of roots of a quadratic function is directly related to the sign of the discriminant in the quadratic equation.