Finding variable in quadratic function through minimum point

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vedd21

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Find the value of h if the quadratic function f(x)=2x^2 +2hx -(h+1) has a minimum value of -5.
(Answer = 2 and -4)

How do I solve this question?I'd appreciate any solution other than trial and error.I just want to know if there is any formula/more appropriate method to solve this equation.Ty
 
vedd21 said:
Find the value of h if the quadratic function f(x)=2x^2 +2hx -(h+1) has a minimum value of -5.
(Answer = 2 and -4)

How do I solve this question?I'd appreciate any solution other than trial and error.I just want to know if there is any formula/more appropriate method to solve this equation.Ty
Click to expand...
There are several ways to calculate that. What
methods have you been taught?
 
vedd21 said:
Find the value of h if the quadratic function f(x)=2x^2 +2hx -(h+1) has a minimum value of -5.
(Answer = 2 and -4)

How do I solve this question?I'd appreciate any solution other than trial and error.I just want to know if there is any formula/more appropriate method to solve this equation.Ty
Click to expand...
One approach is to use the fact that the vertex of `y = ax^2 + bx + c` is located at `x = -b/(2a)`. Plug in what you know, and solve for h.
 
Another is to use the fact that the derivative is 0 at a minimum. The derivative of f(x)= 2x^2+ 2hx- (h+1) is 4x+ 2h. That is 0 when 4x+ 2h= 0 or x= -h/2. f(-h/2)= 2(h^2/4)+ 2h(-h/2)+ h+ 1= h^2/2- h^2+ h+ 1= -h^2/2+ h+ 1= 5. Solve that quadratic equation for h.
 
HallsofIvy said:
Another is to use the fact that the derivative is 0 at a minimum. The derivative of f(x)= 2x^2+ 2hx- (h+1) is 4x+ 2h. That is 0 when 4x+ 2h= 0 or x= -h/2. f(-h/2)= 2(h^2/4)+ 2h(-h/2)+ h+ 1= h^2/2- h^2+ h+ 1= -h^2/2+ h+ 1= 5. Solve that quadratic equation for h.
Click to expand...
This is posted in Beginning Algebra.
 
Subhotosh Khan said:
And we all know that the posters ALWAYS choose the "forum" very carefully matching their "expected" skill level!!
Click to expand...
No, but we have no choice but to believe that they posted in the correct area unless it is obvious they did not.
 
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