How do you sketch the graph of y=0.5(x-2)^2-2 and describe the transformation?

1 Answer
Jayden N.
Aug 10, 2017
  • The y-intercept is at the origin, (0, 0).
  • The x-intercepts are
    (0,0) and (4,0).
  • The turning point is (2, -2).
    graph{ y = 0.5(x-2)^2-2 [-10, 10, -5, 5]}

The graph has shifted 2 units to the right and 2 units down compared to y=x^2. The graph is also wider than y=x^2 since the value of a in the equation is 1/2.

Explanation:

Since the equation is in the form of y = a(x - h)^2 + k, the graph's turning point shifts 2 units to the right, and 2 units down. Therefore, the turning point is (2, -2). The next step is to find the intercepts.

Recall that:

  • To find the y-intercept, let x-0.
  • To find the x-intercepts, let y=0, factorise the equation, and solve for x.

Let x = 0:
y = 0.5((0)-2)^2 - 2
y = 0

The y-intercept is at the origin, (0, 0).

Let y = 0:

0 = 0.5(x-2)^2 - 2

0 = 0.5((x-2)^2 - 4)

0 = (x-2)^2 - 4

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

0 = x(x-4)

x - 4 = 0, x = 0
x = 4, x = 0

Therefore, x-intercepts are (0,0) and (4,0).

Now that you have all the information, let's graph it along with y=x^2.

graph{y=0.5(x-2)^2 - 2 [-10, 10, -5, 5]}

As you can see, the graph has shifted 2 units to the right and 2 units down compared to y=x^2. The graph is also wider than y=x^2 since the value of a in the equation is 1/2.

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