Question

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

Answer 1

• 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`.

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

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

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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)`.

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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`.