How do you graph #y = x^2 - x + 5#?

Question

Isaiah Anthony · Accepted Answer

To graph quadratics, it is usually easier to convert to vertex form, #y = a(x - p)^2 + q

We can convert to vertex form by completing the square. #y = x^2 - x + 5# #y = 1(x^2 - x + m) + 5# #m = (b/2)^2# #m = (-1/2)^2# #m = 1/4# #y = 1(x^2 - x + 1/4 - 1/4) + 5# #y = 1(x - 1/2)^2 - 1/4+ 5# #y = (x - 1/2)^2 + 19/4# Thus, we have our converted equation. Now, let's identify what's what. The vertex is at #(p, q) -> (1/2, 19/4)#. The vertex is the minimum point in the function. The value of the parameter a in #y = a(x - p)^2 + q# is 1. Since it's positive, the parabola opens upwards. The y intercept is at (0, 5). I often recommend finding at least 3 points other than the vertex to make graphing easier. Plug in values for x to find the corresponding y value: #(2, 7); (3, 11); (-1,7)# Now, we have enough information to graph: graph{y = x^2 - x + 5 [-9.375, 10.625, -0.04, 9.96]} Practice exercises: Graph the function #y = 1/2x^2 + 3x - 4# Hopefully this helps!

Ethan Adkins · Answer

undefined To graph the function y = x^2 - x + 5, you can follow these steps:

1. Find the vertex of the parabola using the formula x = -b / (2a), where a is the coefficient of x^2, and b is the coefficient of x.

2. Substitute the x-coordinate of the vertex into the equation to find the y-coordinate.

3. Plot the vertex on the coordinate plane.

4. Find the y-intercept by setting x = 0 and solving for y.

5. Find the x-intercepts by setting y = 0 and solving for x.

6. Plot the intercepts on the graph.

7. Sketch the parabola passing through the vertex and the intercepts.