Fill in the blanks:
find values of a, band cso that the graph of the polynomial p(x)=ax2+bx+cpasses through the point (−1,2)and has a horizontal tangent at (11,−9).
(do not convert fractions to decimals in your answers, enter them exactly e. g. 11/27, 1/3 etc.) (hint: form a linear system based on the assumptions and solve it using your favourite method. see your lectures notes on applications of linear systems.)
a=
b=
c=

Hello,

y=ax²+bx+c

Method 1:

The parabola is passing through (-1,2)
==>2=a-b+c (1)
The parabola has a horizontal tangent  at(11,-9)

y'=2ax+b
==>0=2a*11+b ==> b=-22a (2)
(1) and (2)==>a+22a+c=2 ==>23a+c=2 (3)

The parabola is passing through (11,-9)
-9=a*121+11*b+c (4)
(4) and (2)==> a*121+11*(-22a)+c=-9 ==>-121a+c=-9 (5)

(3)-(5)==>144 a=11==>a=11/144
(2)==>b=-22*11/144==>b=-242/144
(3)==>c=2-23*11/144==>c=35/144

Method 2

The vertex is (11,-9) and the parabola is passing through (-1,2)

y=k(x-11)²-9
2=k*(-12)²-9==>k=11/144
The equation is : y=11/144(x-11)²-9 ==>y=11/142x²-242/144x+35/144

It equals 1373.5 that’s the solution