SOLUTION: 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
Algebra.Com
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
RELATED QUESTIONS
i need some help because im having trouble with this problem
4x^2=36
thank you for (answered by jim_thompson5910)
I am having trouble with the following question, please help.
{{{sqrt(16x-4)= 4x-2}}}... (answered by Theo)
I am having a hard time with this problem. Can someone help me please with the steps on... (answered by checkley77)
(-16x^3+4x^2+16x+3) divided by (4x+3)
I am having the hardest time with this problem. (answered by Fombitz)
Hi I am having trouble with this problem: The directions say to multiply
5x(-3x^4+4x)... (answered by arunpaul)
I am having trouble with multiplying rational expressions
The problem is... 6a over 2b... (answered by jim_thompson5910)
How can I "FACTOR COMPLETELY" this question. I am having problem understanding this... (answered by mathslover)
Hi! I have a problem that asks: "Find the focus and directrix of the parabola from the... (answered by lwsshak3)
I have a graph with numbers 10,30,60,100,150,210,280,360 and so on. I am having trouble... (answered by venugopalramana)