Find the equation of the quadratic function with zeros 10 and 12 and a vertex at (11, -2).

Question

misty1212 · Answer

HI!!

misty1212 · Answer

you have actually too much information here but we can do it anyways

misty1212 · Answer

method one is to write it in factored form as 
$$y=a(x-10)(x-12)$$then put $x=11$ set the answer equal to $-2$ and solve for $a$

misty1212 · Answer

in other words, solve 
$$y=a(11-10)(11-12)=-2$$

misty1212 · Answer

the other way is to write it as 
$$y=a(x-11)^2-2$$ and solve for $x$ by plugging in $x=10$ and setting it equal to zero

misty1212 · Answer

in other words, solve 
$$a(10-11)^2-2=0$$

anonymous · Answer

My bad haha didnt mean to add so much. I just got quit confused. So thats what i have so solve?

misty1212 · Answer

yes you need $a$

misty1212 · Answer

if you do it right, either method, you will find $a=1$

anonymous · Answer

Yes. Thank you!

whpalmer4 · Answer

Unless the problem has been edited, there is not too much information, but rather exactly as much as is needed.  There are an infinite number of quadratics with those two zeroes, so you must completely specify another point to uniquely determine one.  That's what we have, no?

whpalmer4 · Answer

Also, you better try plugging in (11,-2) in the equation to make sure of the answer before concluding that you have the correct value for $a$!

whpalmer4 · Answer

@misty1212 I'm afraid you have supplied the wrong answer.

If, as you say, $$y=1(x-10)(x-12)$$is the correct answer, what happens when we evaluate it at $x=11$?

$$y = 1(11-10)(11-12) = 1(1)(-1) = -1$$But the curve is supposed to pass through $(11,-2)$ not $(11,-1)$, so this is incorrect.

whpalmer4 · Answer

Also, your second method is incorrect, in most cases.  Let's suppose that instead of zeros at $x=10$ and $x=12$, we have zeros at $x = 10$ and $x = 15$, with the curve still passing through $(11,-2)$.  By your procedure, I should be able to find the value of $a$ by solving $$y = a(x-11)^2-2 = 0$$But that's going to give the same answer for $a$ as we had with the other set of zeros, and clearly, this is a different curve, right?  Therefore, it is going to produce at least one wrong answer.  There is an important assumption you are leaving unstated here!

misty1212 · Answer

@whydoihavetosignup1 no there is definitely too much information

misty1212 · Answer

it is overdetermined
if you have one zero and the vertex you can find the quadratic

whydoihavetosignup1 · Answer

what?