Question

example 4 solve a linear-quadratic system by graphing
production a software developer determines that her company can model their revenue r from a given product in hundreds of thousands of dollars given the unit price of the product x in dollars with the function $r = -0.1x^2 + 4x$. create a linear-quadratic system and solve it graphically to determine the price for which the company will earn $4.2 million.
step 1 create a linear-quadratic system.
the first equation is the given revenue model $r = $______.
the line that represents revenue of $4.2 million is $r = $______.
so, the linear-quadratic system is:
$r = $______ (1)
$r = 42$ (2)
step 2 graph the system.
graph: revenue ($100,000) on y-axis, price ($) on x-axis, with a parabola and a horizontal line
step 3 determine the solutions.
the graphs of the functions do not intersect at any point, so the system has ______ solutions.

Explanation:

Step1: Completar la primera ecuación

La primera ecuación es el modelo de ingresos dado, \( R = -0.1x^2 + 4x \).

Step2: Completar la segunda ecuación

El ingreso de \$4.2 millones se convierte a centenares de miles de dólares: \( 4.2 \) millones \( = 42 \) centenares de miles, así que la línea es \( R = 42 \).

Step3: Analizar la intersección

Al graficar \( R = -0.1x^2 + 4x \) (una parábola) y \( R = 42 \) (una línea horizontal), observamos que la parábola no corta a la línea \( R = 42 \) (la parábola tiene un vértice en \( x = 20 \), \( R = 40 \), que es menor que 42). Entonces, el sistema no tiene soluciones.

Answer:

Paso 1:

La primera ecuación es \( R = -0.1x^2 + 4x \).
La línea es \( R = 42 \).
Sistema: \( R = -0.1x^2 + 4x \) (1) y \( R = 42 \) (2).

Paso 3:

El sistema tiene \(\boldsymbol{ninguna}\) (o \(0\)) solución.