Question 282084: I am having trouble with this problem thank you.
Find the ordered pair for the vertex.
y=4x^2-16x+20
I've tried using the quadratic equation and my answer was 2 and 6.
I've also tried factoring the polynomial like this:
4(x^2-4x+5) = (x-5)(x+4) and could not get the last digit in equation to be positive. Please help I am really struggling with this.
Do I need to use the formula: -b/2a?
Thank you in advance for all your help.
Marlo
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Find the ordered pair for the vertex.
I've tried using the quadratic equation and my answer was 2 and 6. Wrong!
I've also tried factoring the polynomial like this:
4(x^2-4x+5) = (x-5)(x+4) and could not get the last digit in equation to be positive. Wrong again! This little monster does not factor.
Please help I am really struggling with this. Right!
Do I need to use the formula: -b/2a? Absolutely!
The  coordinate of the vertex is found by:
and the  coordinate of the vertex is found by:
 = f\left(\frac{-b}{2a}\right))
So once you have calculated  , substitute that value into your function and do the arithmetic:
^2\ -\ 16\left(x_v\right)\ +\ 20)
Now, here is why your quadratic formula attempt was wrong. In the first place, the only relationship that the quadratic formula has to the vertex is that values given by the properly completed arithmetic using the formula will be symmetrically positioned with regards to the vertex. In the second place, with this particular quadratic function, you could not have obtained real number solutions -- the graph never crosses the -axis. Had you performed the arithmetic properly for the quadratic formula, you could have just taken the average of the two roots to determine the -coordinate of the vertex.
\ \pm\ \sqrt{(-16)^2\ -\ (4)(4)(20)}}{2(4)}\ =\ \frac{16\ \pm\ \sqrt{-64}}{8}\ =\ \frac{16\ \pm\ \sqrt{64}\sqrt{-1}}{8}\ =\ \frac{16\ \pm\ 8i}{8}\ =\ 2\ \pm\ i)
Note that the average of the two roots is:
\ +\ (2\ -\ i)}{2}\ =\ 2)
Which is the result you should have gotten when you calculated .
Since we know that for any polynomial function, if  is a root of \ =\ 0) , then  must be a factor of ) , having shown that the roots are a conjugate pair of complex numbers, we can say for certain the the function does not factor over the reals. Had the function been factorable, you could have determined the roots, averaged them, and gotten to the same place.
Super Deluxe Double-Plus Extra Credit
Why does averaging the roots get you the -coordinate of the vertex? Hint: Calculate
John
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