Which graph below solves the following system of equations correctly? y = three over four times x squared minus 3 y = negative three over four times x squared plus 3 A) two quadratic graphs opening up. They intersect at 0 and negative 3. B)one quadratic graph opening up and one quadratic graph facing down. They intersect at 0, 3. C) quadratic graph opening up and quadratic graph opening down. They intersect at 0, negative 3. D) two parabolas one facing down with a vertex at 0, 3 and one facing up with a vertex at 0, negative 3

Answer:D) two parabolas one facing down with a vertex at 0, 3 and one facing up with a vertex at 0, negative 3Step-by-step explanation:First of all, let's rewrite the equations in a mathematical language:y = three over four times x squared minus 3:[tex]y=\frac{3}{4}x^2-3[/tex]Since the leading coefficient, the number that accompanies [tex]x^2[/tex] is positive, that is, its value is 3/4, then the parabola opens upward. On the other hand, the vertex can be found as:[tex](h,k)=\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right) \\ \\ a=3/4 \\ b=0 \\ c=-3 \\ \\ h=-\frac{0}{2(3/4)}=0 \\ \\ k=f(0)=\frac{3}{4}(0)^2-3=-3 \\ \\ \\ \boxed{Vertex \rightarrow (h,k)=0,-3}[/tex]y = negative three over four times x squared plus 3:[tex]y=-\frac{3}{4}x^2+3[/tex]Since the leading coefficient is negative, that is, its value is -3/4, then the parabola opens downward. Similarly the vertex can be found as:[tex](h,k)=\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right) \\ \\ a=-3/4 \\ b=0 \\ c=3 \\ \\ h=-\frac{0}{2(-3/4)}=0 \\ \\ k=f(0)=-\frac{3}{4}(0)^2+3=3 \\ \\ \\ \boxed{Vertex \rightarrow (h,k)=0,3}[/tex]Both graph are shown below and you can see that the conclusion of our problem is correct.