What is the vertex form of #y=7x^2 +3x + 5 #?
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The vertex form of the quadratic equation (y = 7x^2 + 3x + 5) is (y = 7(x + \frac{3}{14})^2 + \frac{191}{28}).
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The vertex form of the quadratic function (y = 7x^2 + 3x + 5) is (y = 7(x - h)^2 + k), where ((h, k)) represents the coordinates of the vertex.
To find the vertex form:
- Complete the square for the quadratic expression.
- Rewrite the expression in the form (y = a(x - h)^2 + k), where (a), (h), and (k) are constants.
Completing the square: [y = 7x^2 + 3x + 5] [y = 7(x^2 + \frac{3}{7}x) + 5]
To complete the square, add and subtract ((\frac{3}{14})^2 = \frac{9}{196}): [y = 7(x^2 + \frac{3}{7}x + \frac{9}{196} - \frac{9}{196}) + 5] [y = 7\left(x + \frac{3}{14}\right)^2 - \frac{9}{28} + 5] [y = 7\left(x + \frac{3}{14}\right)^2 - \frac{9}{28} + \frac{140}{28}] [y = 7\left(x + \frac{3}{14}\right)^2 + \frac{131}{28}]
Therefore, the vertex form of (y = 7x^2 + 3x + 5) is (y = 7\left(x + \frac{3}{14}\right)^2 + \frac{131}{28}).
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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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