How do you write #y=x^2-6x+8# in vertex form and identify the vertex, y intercept and x intercept?
When x = 0 --> y = 8, the y-intercept
Eight factor pairs are composed as follows: (1, 8)(2, 4). The sum equals -b. Therefore, the two real roots (x-intercepts) are 2 and 4.
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To write ( y = x^2 - 6x + 8 ) in vertex form, complete the square: ( y = x^2 - 6x + 8 ) ( y = (x^2 - 6x + 9) - 9 + 8 ) ( y = (x - 3)^2 - 1 )
Vertex: (3, -1) y-intercept: (0, 8) x-intercepts: To find x-intercepts, set y = 0 and solve the quadratic equation. ( x^2 - 6x + 8 = 0 ) gives two x-intercepts, which can be calculated using the quadratic formula or factoring.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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