Question

Evaluate the given function. The values of the independent variable are approximate. f(x)= 6x^2 - 3x, find f(5.24) and f(-8.36) (Do not round until the final answer. Then round to one decimal place as needed.)

Answer 1

In order to evaluate the function at the given values of x, let's use these values of x in the function and calculate its value.

So, for f(5.24), we have x = 5.24:

$\begin{gathered} f(x)=6x^2-3x \\ f(5.24)=6\cdot5.24^2-3\cdot5.24 \\ f(5.24)=6\cdot27.4576-15.72 \\ f(5.24)=164.7456-15.72 \\ f(5.24)=149.0256 \end{gathered}$

Rounding to one decimal place, we have f(5.24) = 149.0.

Now, for f(-8.36), we have:

$\begin{gathered} f(-8.36)=6\cdot(-8.36)^2-3\cdot(-8.36) \\ f(-8.36)=6\cdot69.8896+25.08 \\ f(-8.36)=419.3376+25.08 \\ f(-8.36)=444.4176 \end{gathered}$

Rounding to one decimal place, we have f(-8.36) = 444.4.