Question

Which graph has the x intercepts of -2 and 1? y=x^2-2x+1 y=x^2+x-2 y=2x-1 y=x^2-x-2

Answer 1

To find the graph with x-intercepts of -2 and 1, we can use the fact that x-intercepts occur where \( y = 0 \). This means we need to set the equations provided equal to zero and solve for \( x \).

1. **For \( y = x^2 - 2x + 1 \)**:
\[
x^2 - 2x + 1 = 0 \quad \Rightarrow \quad (x - 1)^2 = 0 \quad \Rightarrow \quad x = 1
\]
This has only one x-intercept at \( x = 1 \).

2. **For \( y = x^2 + x - 2 \)**:
\[
x^2 + x - 2 = 0
\]
We can factor this as:
\[
(x + 2)(x - 1) = 0 \quad \Rightarrow \quad x = -2 \text{ or } x = 1
\]
This has x-intercepts at \( x = -2 \) and \( x = 1 \).

3. **For \( y = 2x - 1 \)**:
\[
2x - 1 = 0 \quad \Rightarrow \quad 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2}
\]
This has only one x-intercept at \( x = \frac{1}{2} \).

4. **For \( y = x^2 - x - 2 \)**:
\[
x^2 - x - 2 = 0
\]
We can factor this as:
\[
(x - 2)(x + 1) = 0 \quad \Rightarrow \quad x = 2 \text{ or } x = -1
\]
This has x-intercepts at \( x = 2 \) and \( x = -1 \).

The only equation that has the x-intercepts -2 and 1 is:
\[
y = x^2 + x - 2
\]

So, the correct answer is \( y = x^2 + x - 2 \).