# How do you find the vertex and intercepts for #4x - y² + 6y - 1 = 0#?

The structure of this is such that it implies

Write as:

Divide both sides by 4

Write as:

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Set

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Set

Where

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'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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To find the vertex and intercepts for the equation (4x - y^2 + 6y - 1 = 0):

- To find the x-intercepts, set (y = 0) and solve for (x).
- To find the y-intercepts, set (x = 0) and solve for (y).
- To find the vertex, use the formula for the vertex of a parabola: (x = -\frac{b}{2a}) and (y = f(x)), where (a) and (b) are coefficients of the quadratic terms.

First, rewrite the equation in standard form: [y^2 - 6y = 4x - 1]

Then, complete the square for the (y) terms: [y^2 - 6y + 9 = 4x - 1 + 9] [(y - 3)^2 = 4x + 8]

Now, the equation

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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.

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