Question

ill medal and its multiple choice .

During a car's lifetime its value will slowly decrease and then, once it becomes a classic, will slowly increase. If the value, V(t) is a function of time, then during part of a particular car's life V(t) = 15.4t2 - 785t + 12000. According to this model, what will be the car's lowest approximate value?


$1500


$2000


$2500


$3000

Answer 1

\[V(t) = 15.4t^2 - 785t + 12000\] you want the vertex of this parabola

Answer 2

the second coordinate of the vertex really
the first coordinate is \(-\frac{b}{2a}\) and the second coordinate is what you get when you replace \(x\) by that number

Answer 3

in your example
\[a=15.4,b=-785\] find
\[\frac{785}{2\times 15.4}\] first

Answer 4

mmmm, I would first factor out of the coefficient of x^2, then be adding inside the parenthesis.
Such.
\(\large\color{black}{ y=c_1(x^2+c_2x)+c_3 }\)
\(\large\color{black}{ y=c_1(x^2+c_2x\color{red}{+(c_2/2)^2-(c_2/2)^2})+c_3 }\)
\(\large\color{black}{ y=c_1(x^2+c_2x\color{red}{+(c_2/2)^2})+c_3 \color{red}{-\color{black}{c_1}(c_2/2)^2} }\)

Answer 5

\(\large\color{black}{ c_1,~~c_2,~~c_3 }\) are just 3 different numbers.

Answer 6

thats confusing.

Answer 7

wow that is an awful lot of work to compute the first coordinate of the vertex \(-\frac{b}{2a}\)
i guess you could also take the derivative, set it equal to zero and solve
probably get \(-\frac{b}{2a}\) also

Answer 8

no, taking the derivative might be something that the user haven't learned yet.
Also this actually isn't hard.

Answer 9

maybe the way I put it with c1 and c2, but we can do an example, if you want.

Answer 10

yes pleas.

Answer 11

is it $2000 @SolomonZelman

Answer 12

there may be a typo in your question
the minimum value of that parabola is somewhere around \(-8803.65\)

Answer 13

oops no you are right, thanks @freckles

Answer 14

http://www.wolframalpha.com/input/?i=15.4t^2+-+785t+%2B+12000 \[1996.35\]