To graph the equation y = x^2 – 3x + 2, we can follow a few steps:
1. Plotting the vertex:
The vertex of the parabola is given by the formula x = -b/2a. In this case, a = 1 and b = -3. Plugging these values into the formula, we get x = -(-3)/(2*1) = 3/2. So the x-coordinate of the vertex is 3/2.
To find the y-coordinate of the vertex, substitute the x-coordinate into the equation. In this case, substituting x = 3/2 into y = x^2 – 3x + 2, we get y = (3/2)^2 – 3(3/2) + 2 = 1/4 – 9/2 + 2 = -15/4.
So the vertex is at (3/2, -15/4).
2. Finding the y-intercept:
To find the y-intercept, substitute x = 0 into the equation. In this case, y = (0)^2 – 3(0) + 2 = 2. So the y-intercept is at (0, 2).
3. Finding the x-intercepts:
To find the x-intercepts, set y = 0 and solve for x. In this case, 0 = x^2 – 3x + 2 can be factored as (x-1)(x-2) = 0. So the x-intercepts are at x = 1 and x = 2.
4. Plotting the points and drawing the graph:
Using the information we found, we can now plot the vertex (3/2, -15/4), the y-intercept (0, 2), and the x-intercepts (1, 0) and (2, 0) on the coordinate grid. Then, we can draw a smooth curve passing through these points.
The graph of the equation y = x^2 – 3x + 2 should look like a U-shaped curve opening upwards, with the vertex at (3/2, -15/4).
I hope this helps! Let me know if you have any further questions.