Lindsey S.

asked • 11/27/15

For the following, determine, without graphing, whether the given quadratic function has a maximum value or a minimum value, and then find that value.

f(x)=-4x-5x2

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Mark M. answered • 11/27/15

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f(x) = -5x2 - 4x                complete the square
f(x) = -5(x2 - (4/5)x
f(x) = -5(x2 - (4/5)x + (4/10)2) + 5(16/100)
f(x) = -5(x2 - (4/10)x)2 + 4/5
 
Downward opening parabola with vertex (maximum) at (-2/5, 4/5)
 
Another method:
 
f(x) = -5x2 - 4x            max/min at -b/2a, -2/5
 
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David W.

Just to elaborate on Mark M's perceptive, precise answer.
 
From Wikipedia:
"A univariate quadratic function can be expressed in three formats:
  f(x) = ax2 + bx +c   is called the standard form,
  f(x) = (x-x1)(x-x2)    is called the factored form, where x1 and x2 are the roots of the quadratic function and the solutions of the corresponding quadratic equation.
  f(x) = a(x-h)2 + k    is called the vertex form, where h and k are the x and y coordinates of the vertex, respectively."
 
So, completing the square puts the given standard form of the equation into the vertex form of the equation of a parabola and immediately identifies the vertex at (h,k).  Pretty perceptive!
 
Now, to check that concise answer, the values of x<(-2/5) should give f(x)<4/5 and the values of x>(-2/5) should give f(x)<(4/5) [since the parabola opens downward).  The value of x=(-2/5) is the vertex that gives the maximum (the vertex) of f(-2/5)=(4/5).
 
(note: Sometimes I'm verbose, but it's usually on purpose to help students understand.)
 
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11/27/15

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