Use the intermediate value theorem to show that the polynomial function has a real zero between the numbers given \[ f(x)=4 x^{2}-2 x-3 ; 1 \text { and } 2 \] What is $f(1) ? \square$ (Simplify your answer Type an integer or a fraction ) What is $1(2)$ ? $\square$ (Simplify your answer. Type an integer or a fraction.) Choose the correct statement below that explains why the given polynomial has a zero between 1 and 2 , according to the intermediate value theorem A. Since $f(1)$ and $t(2)$ are opposite in sign, there exists at least one zero between $t$ and 2 B. Since $f(2)$ is greater than $f(1)$, the function is increasing and so there must be one roal zero between 1 and 2

Step 1 ：Calculate the value of the function at the given points, 1 and 2. This will help us determine if there is a sign change between these two points. If there is a sign change, then by the Intermediate Value Theorem, there must be a root between these two points.

Step 2 ：Calculate \(f(1)\) and \(f(2)\) for the function \(f(x)=4x^{2}-2x-3\).

Step 3 ：\(f(1) = 4(1)^{2}-2(1)-3 = -1\)

Step 4 ：\(f(2) = 4(2)^{2}-2(2)-3 = 9\)

Step 5 ：The function value at \(x=1\) is -1 and at \(x=2\) is 9. Since the function value changes sign between \(x=1\) and \(x=2\), there must be a root between these two points according to the Intermediate Value Theorem.

Step 6 ：\(\boxed{\text{The correct statement is A. Since } f(1) \text{ and } f(2) \text{ are opposite in sign, there exists at least one zero between 1 and 2.}}\)