Question

19. Suppose that the equation V = 21.1x2 – 379.3x + 3,000 represents the
value of a car from 1955 to 2000. What year did the car have the least
value? (Let x = 0 in 1995)

Answer 1

Using the vertex of the quadratic equation, it is found that the car had the least value in the year of 1964.

What is the vertex of a quadratic equation?

A quadratic equation is modeled by:

$y = ax^2 + bx + c$

The vertex is given by:

$(x_v, y_v)$

In which:

$x_v = -\frac{b}{2a}$

$y_v = -\frac{b^2 - 4ac}{4a}$

Considering the coefficient a, we have that:

• If a < 0, the vertex is a maximum point.

• If a > 0, the vertex is a minimum point.

In this problem, the equation is:

$V(x) = 21.1x^2 - 379.3x + 3000$

Hence the coefficients are a = 21.1, b = -379.3, c = 3000, and the minimum value is in x years after 1955, considering that:

$x_v = -\frac{b}{2a} = \frac{379.3}{42.2} \approx 9$

The car had the least value in the year of 1964.

More can be learned about the vertex of a quadratic equation at https://brainly.com/question/13773803