v(x)=-x^2+14x-57

Question

maximum and minimum value , find that value 

 

Answer 1

Since this says quadratics and not Calculus, I will give you the method for Algebra.

 

The equation is a quadratic because the highest power, or exponent, is 2.  There are 2 different ways to work this problem with Algebra.    I will list those  ways and show you how I would choose...

 

Method 1.  Make a chart of values, plot points and graph.  If you find enough points, you should be able to see the shape of the graph and narrow your search for the minimum and maximum...  If you do the graphing on a calculator, this method is even easier.

 

Method 2.  Note that the equation is in standard form...i.e. the Powers are descending.

 

The coefficient in front of x^2 is -1.  Call that A.  A=-1

The coefficient in front of the x-term is 14.  so B=14

The constant term is -57, but we don't need it here.

 

There is a handy-dandy formula for finding the vertex.  The x-coordinate of the vertex is -B/2A.

 

Substituting our A and B into that formula, we get -(14)/2(-1) = 7   

 

The maximum or minimum point of the graph is the y-coordinate of the vertex.  So substitute 7 into your function for x.  You should get V(7) = -(7)^2 + 14*(7) - 57.   Please note that the -1 is not part of the square.  Just square what is in the parentheses.

 

Since this is a downward opening parabola (A is negative), you will have a maximum value.  Remember that the maximum value is v(7) and the value x = 7 is WHERE the maximum occurs.

 

Answer 2

CORRECTION

 

Given a function f(x) = ax² + bx + c, the max/min occur where x = -b/2a

 

We have -x² + 14x - 57, so a = -1 and b = 14

 

The max/min occurs where 

 

x = -b/2a = -14/[2(-1)] = -14/-2 = 7

 

When x = -b/2a is greater than 0, this represents a minimum

When x = -b/2a is less than 0, this represents a maximum

 

Since 7 > 0, then this point is a minimum