How do you find the vertex of #y= x^2-8x+5#?
The simplest method for finding the vertex of the given polynomial is to convert it into "vertex form" to find the vertex at
The "completion of the square method" is the simplest way to accomplish this conversion.
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Find vertex of f(x) = x^2 - 8x + 5
Ans: Vertex (4, -11)
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To find the vertex of the quadratic function ( y = x^2 - 8x + 5 ), you can use the formula ( x = -\frac{b}{2a} ). In this equation, ( a ) represents the coefficient of the ( x^2 ) term, and ( b ) represents the coefficient of the ( x ) term. Plugging the values ( a = 1 ) and ( b = -8 ) into the formula, you get ( x = -\frac{-8}{2(1)} ), which simplifies to ( x = 4 ). To find the corresponding ( y )-coordinate, substitute ( x = 4 ) into the original function: ( y = (4)^2 - 8(4) + 5 = 16 - 32 + 5 = -11 ). Therefore, the vertex of the function is ( (4, -11) ).
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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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