Find ex and Y to the quadrilateral is a parallelogram.
1 answer:
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Let's first rewrite it in vertex form.
Start by completing the square.
y = x² + 4x + 4 - 4 - 1
y = (x + 2)² - 5
(x + 2)² = y + 5
Now, 4a = 1; a =
So, the vertex form is (x + 2)² = 4()(y + 5)
So, we know that the vertex is at (-2, -5).
Since it's a concave up parabola, we just have to change the y-value to find the focus.
∴ F(-2, -5 +)
F(-2,
)
Start by completing the square.
y = x² + 4x + 4 - 4 - 1
y = (x + 2)² - 5
(x + 2)² = y + 5
Now, 4a = 1; a =
So, the vertex form is (x + 2)² = 4(
So, we know that the vertex is at (-2, -5).
Since it's a concave up parabola, we just have to change the y-value to find the focus.
∴ F(-2, -5 +
F(-2,
