Maximum of Quadratic

Algebra Level 1

Consider the quadratic function

y = a x 2 + b x + c , y=ax^2+bx+c,

where a a , b b and c c are constants. If y y has a maximum value of 54 54 , and the graph of y y passes through the two points ( − 2 , 0 ) (-2, 0) and ( 4 , 0 ) , (4, 0), what is the value of a + b + c ? a+b+c?

54 56 58 60

Ajay Dutta by Ajay Dutta , 24, India

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2 solutions

Alain Chau
Sep 10, 2016

Since the roots are -2 and 4, the parabola attains its maximum at x = ( − 2 + 4 ) / 2 = 1 x=(-2+4)/2=1 . Hence 54 = y = a ( 1 ) 2 + b ( 1 ) + c = a + b + c . 54 = y = a(1)^2 + b(1) + c = a+b+c.

7 Helpful 3 Interesting 3 Brilliant 3 Confused

Nice solution!

Anurag Pandey - 4 years, 9 months ago
Ravindra Sahu
Mar 1, 2014

since the quad. inequality has a max. value , the graph of it is an inverted parabola and maximum value occurs at x = -b/2a. and maximum value is given by; y_max = c-sqr.(b)/4*sqr(a) = 54 ( given). ............(1) also by satisfying the given 2 points in parabolas eqn . we get other 2 eqns as ; 4a-2b +c = 0 ...........(2) 16a+4b +c=0 .....(3) solving the three eqns we get the relation in a,b,c as : c= -8a ; & b=-2a and a = -6. So it gives us ; a+b+c = 54.(ans.)

Hoping it is clear to all....

3 Helpful 0 Interesting 1 Brilliant 3 Confused

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