SOLUTION: equation V=40x^2-500x+5000 describes he value of a car from 1960 to 2012. What year did the car have the least value? (x=0 is 1960) 1967 1966 1968 1965

Question 824936: equation V=40x^2-500x+5000 describes he value of a car from 1960 to 2012. What year did the car have the least value? (x=0 is 1960)
1967
1966
1968
1965
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
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v = 40x^2 - 500x + 5000
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the above quadratic equation is in standard form, with a=40, b=-500, and c=5000
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
40 -500 5000
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic vertex is a minimum at: ( x= 6.25, v= 3437.5 )
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answer:
the car's value was lowest during the first quarter of the year 1966
at a value of 3437.50 (monetary units)
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Solve quadratic equations, quadratic formula:
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Solve systems of linear equations up to 6-equations 6-variables:
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