https://file.aiquickdraw.com/mathbot/file/1729874889227dr8261i2.jpg

The problem in the image asks to find a quadratic polynomial \( P(x) \) based on the following conditions:

1. \( P(x) \) can be positive or zero for all real numbers, implying that it has no real roots (the parabola opens upwards).
2. \( P(1) = 0 \), meaning that \( x = 1 \) is a root of the polynomial.
3. \( P(2) = 2 \), which gives another point on the polynomial.

### Solution:

A general quadratic polynomial can be written as:

\[
P(x) = a(x - 1)^2
\]

Since \( P(x) \) must be non-negative for all real numbers, the quadratic opens upwards, and the root at \( x = 1 \) is a double root (this satisfies \( P(1) = 0 \)).

Now, using the second condition \( P(2) = 2 \):

\[
P(2) = a(2 - 1)^2 = 2
\]
\[
a(1)^2 = 2
\]
\[
a = 2
\]

Thus, the quadratic polynomial is:

\[
P(x) = 2(x - 1)^2
\]

### Final Answer:
\[
P(x) = 2(x - 1)^2
\]

Would you like further details on this solution, or any additional help?

Here are five related questions to explore further:
1. What is the general form of a quadratic function?
2. How does the discriminant of a quadratic equation affect the number of roots?
3. How do we derive the vertex form of a quadratic equation?
4. What are some applications of quadratic polynomials in real life?
5. How do transformations affect the graph of a quadratic function?

**Tip:** The vertex form of a quadratic equation is very useful for identifying the minimum or maximum point of a parabola easily!

Find a quadratic polynomial P(x) that can be positive or zero for all real numbers, with P(1) = 0 and P(2) = 2.

Solve a quadratic polynomial P(x) that is non-negative for all real numbers, with conditions P(1)=0 and P(2)=2. Learn step-by-step how to derive the solution.

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