Find the intersection points of the graphs of the equations $y = 2x^2 + 3x - 2$ and $y = 2x - 1$.

1. Problem Description

Find the intersection points of the graphs of the equations $y = 2x^2 + 3x - 2$ and $y = 2x - 1$.

2. Solution Steps

To find the intersection points, we need to solve the system of equations:

$y = 2x^2 + 3x - 2$

$y = 2x - 1$

Since both equations are solved for $y$, we can set them equal to each other:

$2x^2 + 3x - 2 = 2x - 1$

Now, rearrange the equation to form a quadratic equation:

$2x^2 + 3x - 2 - (2x - 1) = 0$

$2x^2 + 3x - 2 - 2x + 1 = 0$

$2x^2 + x - 1 = 0$

We can solve this quadratic equation using the quadratic formula or by factoring. Let's try factoring:

$2x^2 + 2x - x - 1 = 0$

$2x(x + 1) - 1(x + 1) = 0$

$(2x - 1)(x + 1) = 0$

This gives us two possible solutions for $x$:

$2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$

$x + 1 = 0 \implies x = -1$

Now, we need to find the corresponding $y$ values for each $x$ value. We can use the equation $y = 2x - 1$.

For $x = \frac{1}{2}$:

$y = 2(\frac{1}{2}) - 1 = 1 - 1 = 0$

So, one intersection point is $(\frac{1}{2}, 0)$.

For $x = -1$:

$y = 2(-1) - 1 = -2 - 1 = -3$

So, the other intersection point is $(-1, -3)$.

3. Final Answer

The intersection points are $(\frac{1}{2}, 0)$ and $(-1, -3)$.