I was given the problem of f(x)=-x^2+2x-3 and I know that you have to find the vertex by finding -b/2a for the x value but for the y, when I plug the x value back into the equation I only get 0 when the answer says it should be -2, why am I getting this wrong?
Vertex of a quadratic function
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Dr.Peterson
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What did you get for x?
I suspect you may have a sign error in the first term. Try evaluating y again.
I suspect you may have a sign error in the first term. Try evaluating y again.
Dr.Peterson
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Let me show what I think you may have done:
We have a=-1, b=2, c=-3; so x = -b/(2a) = -2/-1 = 1.
So y = -x^2+2x-3 = -(1)^2+2(1)-3. If you thought this meant (-1)^2+2(1)-3, you would get 1+2-3 = 0. But if you do it correctly, you get -1+2-3 = -2.
If I'm right, this is an order of operations issue. When we write -x^2, that means -(x^2), that is, the negative of the square of x, not (-x)^2, the square of the negative of x.
Have I diagnosed the problem correctly? As Jomo said, I could have done that immediately if you had shown more of your thinking.
pka
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Given \(\displaystyle f(x)=ax^2+bx+c\) then the vertex occurs when \(\displaystyle 2ax+b=0\) or \(\displaystyle x=\frac {-b}{2a}\).
In this question \(\displaystyle a=-1~\&~b=2\). Please post what you get now.
HallsofIvy
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Complete the square: \(\displaystyle -x^2+ 2x- 3= -(x^2- 2x)- 3= -(x^2- 2x+ 1- 1)- 3\)\(\displaystyle = -(x^2- 2x+ 1)+ 1- 3= -(x+ 1) ^2- 2\).
Since a square is never negative, That is -2 when x= -1 and is less than -2 otherwise- its vertex is at (-1, -2).
Since a square is never negative, That is -2 when x= -1 and is less than -2 otherwise- its vertex is at (-1, -2).