Find the maximum or minimum value of the function. Approximate to two decimal places.
-$f ( x ) = 1.4 x ^ { 2 } - 1.2 x - 0.9$
A) $0.43$
B) $- 1.16$
C) $- 0.43$
D) $- 0.13$

Question

Quizplus · Accepted Answer

The function $f(x) = 1.4x^2 - 1.2x - 0.9$ is a quadratic function in the form $ax^2 + bx + c$, where $a = 1.4$, $b = -1.2$, and $c = -0.9$. Since $a > 0$, the parabola opens upwards, indicating that the vertex represents the minimum point of the function. The x-coordinate of the vertex is given by $-\frac{b}{2a}$, and the y-coordinate (the minimum value of the function) can be found by substituting this x-value back into the function.The x-coordinate of the vertex is:$$x = -\frac{-1.2}{2(1.4)} = \frac{1.2}{2.8} = \frac{3}{7} \approx 0.43$$Substituting $x = 0.43$ back into the function to find the minimum value:$$f(0.43) = 1.4(0.43)^2 - 1.2(0.43) - 0.9$$Calculating this gives:$$f(0.43) \approx -1.16$$Therefore, the minimum value of the function, approximated to two decimal places, is $-1.16$.

Find the Maximum or Minimum Value of the Function f(x)=1.4x2−1.2x−0.9f ( x ) = 1.4 x ^ { 2 } - 1.2 x - 0.9f(x)=1.4x2−1.2x−0.9

Find the Maximum or Minimum Value of the Function $f ( x ) = 1.4 x ^ { 2 } - 1.2 x - 0.9$