SAT Mathematics Practice Test 4

The minimum value of a quadratic function appears as a constant in the vertex form of its equation, which can be found from the standard form by completing the square. Rewriting $\mathit{f}\left(\mathit{x}\right)=\left(\mathit{x}+6\right)\left(\mathit{x}-4\right)$ in standard form gives $\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^2+2\mathit{x}-24$. Since the coefficient of the linearterm is $2$, the equation for $\mathit{f}\left(\mathit{x}\right)$ can be rewritten in terms of ${\left(\mathit{x}+1\right)}^2$ as follows:

Since the square of a real number is always nonnegative, the vertex form $\mathit{f}\left(\mathit{x}\right)={\left(\mathit{x}+1\right)}^2-25$ shows that the minimum value of f is $-25$ (and occurs at $\mathit{x}=-1$). Therefore, this equivalent form of f shows the minimum value of $\mathit{f}$ as a constant.
Choices A and C are incorrect because they are not equivalent to the given equation for $\mathit{f}$. Choice B is incorrect because the minimum value of $\mathit{f}$, which is $-25$, does not appear as a constant or a coefficient.