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the question is $$y = \sqrt{-(x -3}) + 4$$ i thought the vertex was (3,4) but i was wrong and that it was supposed to be (3,2). Was i right, can anyone help me with these type of question?

MATH ASKER
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  • Did you graph the function? – imranfat Sep 30 '15 at 01:51
  • yes i tried to but as my vertex was wrong i could not..for this assignment i was told not to use the calculator so when i was not using the calculator my vertex was (3,4) but when i checked the calculator to be sure i found out that it was (3,2). Can you tell me what i did wrong or a brief explanation on how to find it? – MATH ASKER Sep 30 '15 at 01:54
  • The answer is (3,4) according to my calculator. The y- coordinate cannot be 2, because that would result in solving the equation $ -2=\sqrt{3-x}$ Can you see why that cannot be done? – imranfat Sep 30 '15 at 01:55
  • So does that mean i was right...so to find the vertex its (h,k) right? – MATH ASKER Sep 30 '15 at 02:00
  • Yes, you were... – imranfat Sep 30 '15 at 02:04

2 Answers2

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Notice, we have $$y=\sqrt{-(x-3)}+4$$ $$y-4=\sqrt{3-x}$$ Squaring both the sides, we get

$$(y-4)^2=3-x$$$$$\implies (y-4)^2=-(x-3)$$ Now, compare the above equation with the standard equation of the parabola: $Y^2=-4AX$ having vertex at the origin $(0, 0)$ & its axis coincident with the x-axis, we get $$X=x-3, \ Y=y-4, A=\frac{1}{4}$$ Then the vertex of the parabola is given as $$(X=0, Y=0)$$ $$(x-3=0, y-4=0)\equiv\color{red}{(3, 4)}$$

Harish Chandra Rajpoot
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We know that a square root equation's vertex is at the point where the part under the square root is $0$ (at which point it stops, because you can't have a real square root of a negative number). Solving, we get $-(x-3)=0\implies x-3=0\implies x=3\implies y=0+4\implies y=4$. Thus, our vertex is $(3,4)$.

thepiercingarrow
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