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What is the y coordinate of the vertex of a parabola #y=x^2-4x+12#?

Answer 1

Hazel Anglin

Vertex is #(2, 8)#

Method Arrange the equation in the form #y - k = 4a(x - h)^2#

where #(h, k)# is your vertex.

Thus, we have

#=>y = x^2 - 4x + 12#

Upon finishing the square,

#=> y = x^2 - 4x + (-4/2)^2 - (-4/2)^2 + 12#

#=> y = (x - 2)^2 - 4 + 12#

#=> y - 8 = (x - 2)^2#

So we get our vertex to be #(2, 8)#

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Answer 2

Kennedy Adams

The y-coordinate of the vertex of the parabola ( y = x^2 - 4x + 12 ) can be found using the formula ( y = ax^2 + bx + c ) with the vertex formula ( x_v = -\frac{b}{2a} ). Substituting the values ( a = 1 ) and ( b = -4 ) into the formula, we get ( x_v = -\frac{-4}{2(1)} = 2 ). Then, substitute ( x_v = 2 ) into the equation ( y = x^2 - 4x + 12 ) to find the y-coordinate. ( y = (2)^2 - 4(2) + 12 = 4 - 8 + 12 = 8 ). Therefore, the y-coordinate of the vertex is 8.

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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.