How many roots does the function $^{y=a(x+p)^2+q}$ have if… a>0, q>0

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.

How many roots does the function $y=a(x+p)^2+q$ have if a>0, q>0?

Explore the solution to the quadratic function y = a(x+p)^2 + q with conditions a > 0 and q > 0. This function represents a parabola that does not intersect the x-axis, and thus has no real roots. Suitable for students in Grades 9-11 learning about quadratic equations and parabolas.

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