SAT Question f123e039: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice D is correct. It’s given that the graph shows a system consisting of a linear and a quadratic equation in the xy-plane. In each of the given choices, the linear equation is in the form $y=mx$, where $m$ is a constant, and the quadratic equation is in the form $y=x^2+c$, where $c$ is a constant. The graph of a linear equation in the form $y=mx$ has a slope of $m$ and a y-intercept at $(0,0)$. Given two points on the line, $(x_1,y_1)$ and $(x_2,y_2)$, the slope can be calculated using the formula $m=\frac{y_2-y_1}{x_2-x_1}$. The graph of the linear equation in this system has a y-intercept at $(0,0)$ and passes through the point $(1,5)$. Substituting the points $(0,0)$ and $(1,5)$ for $(x_1,y_1)$ and $(x_2,y_2)$ in the formula $m=\frac{y_2-y_1}{x_2-x_1}$ yields $m=\frac{5-0}{1-0}$, which is equivalent to $m=\frac{5}{1}$, or $m=5$. It follows that the linear equation is $y=5x$. The graph of a quadratic equation in the form $y=x^2+c$ has a vertex at $(0,c)$. The graph of the quadratic equation in this system has a vertex at $(0,-8)$. It follows that the quadratic equation is $y=x^2+(-8)$, or $y=x^2-8$. Thus, the system of equations represented by the graph is $y=5x$ and $y=x^2-8$.