149k views
2 votes
Find the minimum value of the function f(x) = 0.8x2 + 11.2x + 42 to the
nearest hundredth.

User Tpdi
by
8.2k points

1 Answer

4 votes

Answer:

The minimum value is 2.8. The vertex is (-7, 2.8).

Explanation:

Through the x^2 term we identify this as a quadratic function whose graph opens up. The minimum value of this function occurs at the vertex (h, k). The x-coordinate of the vertex is

-b 11.2

x = --------- which here is x = - ------------ = -7

2a 2(0.8)

We need to calculate the y value of this function at x = -7. This y-value will be the desired minimum value of the function.

Substituting -7 for x in the function f(x) = 0.8x^2 + 11.2x + 42 yields

f(-7) = 0.8(-7)^2 + 11.2(-7) + 42 = 2.8

The minimum value is 2.8. The vertex is (-7, 2.8).

User Hardik Gondalia
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories