Question

or
What is the range of this quadratic function?
y=x2–10x+21

Answer 1

To find the range of the quadratic function \( y = x^2 - 10x + 21 \), we can follow these steps:

1. **Rewrite the quadratic in vertex form**: We can complete the square to rewrite the quadratic function.

Starting with:
\[
y = x^2 - 10x + 21
\]

We complete the square:
\[
y = (x^2 - 10x + 25) - 25 + 21
\]
\[
y = (x - 5)^2 - 4
\]

2. **Identify the vertex**: The expression \( (x - 5)^2 \) is always non-negative, and it achieves its minimum value of 0 when \( x = 5 \).
Therefore, at \( x = 5 \):
\[
y = (5 - 5)^2 - 4 = 0 - 4 = -4
\]

This means the vertex of the parabola is at \( (5, -4) \).

3. **Determine the direction of the parabola**: Since the coefficient of \( x^2 \) is positive (1), the parabola opens upwards.

4. **Find the range**: Since the vertex represents the minimum point of the parabola and the parabola opens upwards, the minimum value of \( y \) is -4, and it can take any value greater than or equal to -4.

Thus, the range of the function is:
\[
\text{Range } y = [-4, \infty)
\]