What are the intercepts of y = 2x^2 - 3x + 1?

The intercepts of \( y = 2x^2 - 3x + 1 \) are the y-intercept at (0, 1) and the x-intercepts at (1, 0) and (0.5, 0).

To find the y-intercept, we set \( x = 0 \) and solve for \( y \). Substituting \( x = 0 \) into the equation \( y = 2x^2 - 3x + 1 \), we get:
\[ y = 2(0)^2 - 3(0) + 1 = 1 \]
So, the y-intercept is at the point (0, 1).

Next, to find the x-intercepts, we set \( y = 0 \) and solve for \( x \). This means we need to solve the quadratic equation:
\[ 2x^2 - 3x + 1 = 0 \]
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2 \), \( b = -3 \), and \( c = 1 \). Substituting these values in, we get:
\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(1)}}{2(2)} \]
\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} \]
\[ x = \frac{3 \pm 1}{4} \]
This gives us two solutions:
\[ x = \frac{3 + 1}{4} = 1 \]
\[ x = \frac{3 - 1}{4} = 0.5 \]
So, the x-intercepts are at the points (1, 0) and (0.5, 0).

In summary, the y-intercept is where the graph crosses the y-axis, and the x-intercepts are where the graph crosses the x-axis. For the quadratic equation \( y = 2x^2 - 3x + 1 \), these intercepts are (0, 1), (1, 0), and (0.5, 0) respectively.