How do you write the quadratic in vertex form given #y=-2x^2+6x+1#?

Answer 1
Standard form# f(x) = - 2x^2 + 6x + 1# Vertex form #f(x) = (x - b/(2a))^2 + f(-b/(2a))#
#-b/2a = -6/-4 = 3/2#
#f(3/2) = -2(9/4) + 6(3/2) + 1 = 11/2#
Vertex form# f(x) = (x - 3/2)^2 + 11/2#
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Answer 2

To write a quadratic equation in vertex form, you can use the formula y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. First, factor out the coefficient of x^2 from the given equation: y = -2(x^2 - 3x) + 1. Complete the square inside the parentheses: y = -2(x^2 - 3x + 9/4 - 9/4) + 1. Rewrite the equation: y = -2(x^2 - 3x + 9/4) + 9/2 + 1. Factor and simplify: y = -2(x - 3/2)^2 + 11/2. The quadratic equation in vertex form is y = -2(x - 3/2)^2 + 11/2.

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Answer from HIX Tutor

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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