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How do you find the vertex of #f(x)=2(x-3)^2+1#?

Answer 1

vertex: #color(green)(""(3,1))#

The general quadratic vertex form is #color(white)("XXX")f(x)=color(orange)m(x-color(red)a)^2+color(blue)b# with vertex at #(color(red)a,color(blue)b)#

The given quadratic #color(white)("XXX")f(x)=color(orange)2(x-color(red)3)^2+color(blue)1# is a specific example of this vertex form with vertex at #(color(red)3,color(blue)1)#

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Answer 2

Genesis Allen

To find the vertex of the function ( f(x) = 2(x-3)^2 + 1 ), use the formula ( x = \frac{-b}{2a} ) where ( a ) and ( b ) are coefficients from the quadratic equation ( ax^2 + bx + c ). In this case, ( a = 2 ) and ( b = -6 ). Plug these values into the formula to find the x-coordinate of the vertex. Then, substitute this x-value into the original function to find the corresponding y-coordinate. Thus, the vertex of the function is ( (3, 1) ).

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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.