SAT Mathematics Practice Test 4

One can find the intersection points of the two graphs by setting the functions $\mathit{f}\left(\mathit{x}\right)$ and $\mathit{g}\left(\mathit{x}\right)$ equal to one another and then solving for $\mathit{x}$. This yields $8{\mathit{x}}^2-2=-8{\mathit{x}}^2+2.$ Adding $8{\mathit{x}}^2$ and $2$ to each side of the equation gives $16{\mathit{x}}^2=4.$ Then dividing each side by $16$ gives $\mathit{x}=\pm \frac{1}{2}$ From the graph, the value of k is the $\mathit{x}-$coordinate of the point of intersection on the positive x-axis. Therefore,$\mathit{k}=\frac{1}{2}$
Alternatively, since $\left(\mathit{k},0\right)$ lies on the graph of both f and g, it follows that $\mathit{f}\left(\mathit{k}\right)=\mathit{g}\left(\mathit{k}\right)=0.$ Thus, evaluating $\mathit{f}\left(\mathit{x}\right)=8{\mathit{x}}^2-2$ at $\mathit{x}=\mathit{k}$ gives $0=8{\mathit{k}}^2-2$. Adding $2$ to each side yields $2=8{\mathit{k}}^2$ and then dividing each side by $8$ gives $\frac{1}{4}={\mathit{k}}^2$.Taking the square root of each side then gives $\mathit{k}=\pm \frac{1}{2}$.From the graph, k is positive, so $\mathit{k}=\frac{1}{2.}$
Choices A, C, and D are incorrect and may be the result of calculation errors in solving for $\mathit{x}$ or $\mathit{k}$