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

Answer 1

Vertex: #(1,2)#

An equation in the form #f(x)=color(green)(m)(x-color(red)(a))^2+color(blue)(b)# is in vertex form with vertex at #(color(red)(a),color(blue)(b))#

#f(x)=-(x-1)^2+2# is equivalent to #f(x)=color(green)(""(-1))(x-color(red)(1))^2+color(blue)(2)# and therefore has a vertex at #(color(red)(1),color(blue)(2))#

Here is a graph of the original equation for verification purposes: graph{-(x-1)^2+2 [-2.507, 3.652, -0.54, 2.537]}

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Answer 2

Jameson Allen

To find the vertex of the quadratic function ( f(x) = -(x-1)^2 + 2 ), you first identify the values of ( h ) and ( k ) in the vertex form ( f(x) = a(x - h)^2 + k ). In this equation, ( h ) represents the horizontal shift of the parabola and ( k ) represents the vertical shift. In the given function, ( h = 1 ) and ( k = 2 ). Therefore, the vertex of the function is at the point ( (1, 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.