G33k
  • G33k
i needed to edit this question :p
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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whpalmer4
  • whpalmer4
Do you know what the vertex is for this parabola?
whpalmer4
  • whpalmer4
If you know the vertex, you can write down vertex form immediately. Otherwise, you can rearrange the equation into vertex form: \[y = a(x-h)^2 + k\]where \((h,k)\) is the vertex and \(a\) is a constant which is the coefficient of \(x^2\) in the formula. Yes, I understand that the problem text does not tell you the vertex explicitly. I am asking if you can look at the equation and figure out the vertex. It is easy if you know how.
whpalmer4
  • whpalmer4
if you have a quadratic formula \[y = ax^2 + bx + c\]the x value of the vertex is given by \[x = -\frac{b}{2a}\]you can then plug that value of \(x\) into the formula and get the y value

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whpalmer4
  • whpalmer4
Try it, what do you get?
whpalmer4
  • whpalmer4
Close. \[y = -x^2 + 12x -4\]\[y = ax^2+bx+c\]\[-x^2 = ax^2\]\[a=-1\]\[12x=bx\]\[b=12\]\[-4=c\] \[x = -\frac{b}{2a} = -\frac{12}{2*(-1)} = 6\]
whpalmer4
  • whpalmer4
so if that is the x coordinate of the vertex, what is the y coordinate?
whpalmer4
  • whpalmer4
no. \[y = -x^2 +12x -4\]\[y = -(6)^2 + 12(6) -4 = \]
whpalmer4
  • whpalmer4
oh come, you can do this. what is 6*6?
whpalmer4
  • whpalmer4
right. so \[-(6)^2 = -(6*6) = -36\] \[y = -(6)^2 + 12(6) -4 = -36 + 12(6) - 4 =\]
whpalmer4
  • whpalmer4
I'm just trying to spell out each step so you can see exactly how to do it. \[12(6) = 12*6 = \]
whpalmer4
  • whpalmer4
right! and I'm sure you don't need my help to know that \[-4 = -4\]:-) So, putting the pieces together, we know that we are trying to find the value of the y-coordinate of the vertex. We know that the x-coordinate is \(x=6\) and that if we plug \(x=6\) into our equation:\[y=-x^2+12x-4\]we will get the y-value. \[y = -(6)^2 + 12(6) -4 \]\[y = -36 + 72 -4 \]\[y = 36-4\]\[y=32\]so our vertex is at (drumroll please...) \[(2,32)\] Still with me?
whpalmer4
  • whpalmer4
Great! As I mentioned previously, if we call the coordinates of our vertex \((h,k)\), then we can write the equation in vertex form: \[y = a(x-h)^2 +k\] where \(a\) is the same value it had when we wrote \[y = ax^2 + bx + c\]
whpalmer4
  • whpalmer4
If our vertex is at \((2,32)\) and our original equation was \[y = -x^2+12x-4\] can you fill in the blanks and write the vertex form equation for me?
whpalmer4
  • whpalmer4
no, it's going some of the sort \[y = 3(x-2)^2 + 27\]except that all the numbers will be different :-)
whpalmer4
  • whpalmer4
(well, okay, the exponent will be the same!)
whpalmer4
  • whpalmer4
we found the vertex (which we call \((h,k)\)) to be \((2,32)\) so \((h,k) = (2,32)\) which is just a compact way of writing \[h=2\]\[k=32\]
whpalmer4
  • whpalmer4
getting closer. but you lost the \(a\) and the parentheses around the (x-3) and the you omitted the exponent. \[y = ax^2 + bx + c\]\[y = -x^2 + 12x -4\]so \(a = -1\) our "template" is \[y = a(x-h)^2 + k\] \[y = -1(x-2)^2 + 32\] or that would usually be written \[y = -(x-2)^2 + 32\] Except I am an idiot and didn't notice that we started using the wrong value of \(x\) back there!!! Grrr....
whpalmer4
  • whpalmer4
Let's regroup. I'll go through all the steps. Original formula: \[y = -x^2 + 12x -4\]Template formula for standard form: \[y = ax^2+bx+c\]so we have\[a=-1\]\[b=12\]\[c=-4\] x-coordinate of vertex at \[x = -\frac{b}{2a} = -\frac{12}{2(-1)} = 6\] y-coordinate of vertex at \[y = -x^2+ 12x-4 = -(6)^2+12(6)-4 = 32\]So vertex is at \((h,k) = (6,32)\) Now to write our final solution in vertex form: \[y = a(x-h)^2 +k\]\[y = (-1)(x-6)^2+32\]\[y = -(x-6)^2 + 32\]
whpalmer4
  • whpalmer4
I'm very sorry I accidentally switched \(x\) values as I wrote down the vertex earlier! Must have added to the confusion... :-(
whpalmer4
  • whpalmer4
There's another way we could have obtained our answer. If you expand our answer by multiplying it out: \[y = -(x-6)^2 + 32\]\[y = -[(x-6)(x-6)] + 32\]\[y = -[x^2-6x-6x-6(-6)]+32\]\[y = -[x^2 - 12x + 36] +32\]\[y = -x^2 + 12x -36 + 32\]\[y = -x^2+12x-4\] That's the same thing, clearly. We could gone the other way by taking \[y = -x^2+12x-4\]and figured out how to factor it to look like our desired \[y = a(x-h)^2+k\]form. I think my approach is easier, especially if you aren't good at factoring or completing the square, but I'll do it the other way if you like, as a demonstration that there are often multiple ways to get the answer.
whpalmer4
  • whpalmer4
First, let me show you a graph of the parabola:
1 Attachment
whpalmer4
  • whpalmer4
The crosshairs on the diagram show \(y = 32\) and \(x=6\), which intersect at our vertex, so that you can see that our parabola really does have a vertex where we thought it should be.
whpalmer4
  • whpalmer4
If we wanted to do this by completing the square: \[y = -x^2 +12x -4\]\[y = -1(x^2 -12x) -4\]When we complete the square, we take \[x^2 + bx\]and find a value that we can add to it so that we can write it as \[(x+\frac{b}{2})^2\] If we expand \[(x+\frac{b}{2})^2 = (x+\frac{b}{2})(x+\frac{b}{2}) = x^2 + \frac{b}{2}x -\frac{b}{2}x +\frac{b}{2}(\frac{b}{2})\]\[=x^2+2(\frac{b}{2})x+\frac{b^2}{4}\]\[=x^2+bx+ \frac{b^2}{4}\] So, our \[x^2-12x\]can be replaced by \[(x-\frac{12}{2})^2 + \frac{(-12)^2}{4} = (x-6)^2 +36\]as long as we subtract \(36\) from the equation as well (to offset the \(36\) we just added by completing the square). And if we do that in our equation: \[y - -x^2 + 12x -4\] \[y = -[(x^2-12x)]-4\]\[y = -[(x-6)^2 - 36 ] -4 \]\[y = -(x-6)^2 +36 -4\]\[y = -(x-6)^2+32\] which is exactly what we obtained doing the problem the other way.

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