Bunuel wrote:
sr2013 wrote:
Hi Bunuel
What is wrong in the following approach
1) Insufficient
The vertex of parabola is (2,-5) --> the vertex is in the IV quadrant: if the parabola is downward it'll have negative y-intercept, but if it's upward then it can have positive as well as negative y-intercept. Not sufficient
I was thinking from option 2 we can find the product of the roots and sum of the roots
product of roots = c/a = -12
and sum of roots = -b/a = 4
hence come up with an equation y = x^2-4x-12
and from the question stem we have that it intercepts at 0,y
Putting x = 0 we get y = -12 (negative) and B alone is sufficient .
How did you get y=x^2-4x-12 from c/a=-12 and -b/a=4? You cannot solve c/a=-12 and -b/a=4 to get unique values of a, b, and c.
For example if a=2, c=-12, and b=-4 you'll get 2x^2-24x-8=0:
Attachment:
MSP2141ga1ih773ii7e3di00002h2000h2fiibia6c.gif
If a=-1, c=12, and b=4 you'll get -x^2+4x+12=0:
Attachment:
MSP24971d0ia65iig79i4gg000037eb4cg662i7adih.gif
Or in other words infinitely many parabolas have x intercepts at -2 and 6. You cannot get unique equation only from that info.
Hi Bunuel
I have tried solving it using standard expressions for parabola
For parabola y= ax2+ bx+ c, standard vertex is located at point (-\frac{b}{2a}, c-\frac{b^2}{4a}).
From a) we know the value of vertex as (2,-5)
by putting the value in standard vertex we can get c=-4
it is also given in the question stem that parabola intersects y axis at (0,y)
from this we can get the value of y as -4. which is sufficient to answer the question.
Please let me know whats wrong with this approach.