Two parabolas from a graph and their difference

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The figure shows two parabolas $R$ and $J$. From the graph you read that $R$ opens upward, has vertex $V_R(-3;1)$ and passes through $(-1;3)$; while $J$ opens downward, meets the $x$-axis at $x=-5$ and $x=1$, and has vertex $V_J(-2;3)$.

• (a) find the vertices and axis intercepts of $R$ and $J$;
• (b) find the equation of $R$;
• (c) find the equation of $J$;
• (d) show that $R-J=\dfrac{5x^2}{6}+\dfrac{13x}{3}+\dfrac{23}{6}$;
• (e) set $R-J=0$, solve, and explain the meaning of the solutions.

Solution

(b) Equation of $R$. From the vertex $V_R(-3;1)$: $R:\;y=a(x+3)^2+1$. Through $(-1;3)$: $a(2)^2+1=3\Rightarrow 4a=2\Rightarrow a=\tfrac12$. Hence $R:\;y=\tfrac12(x+3)^2+1=\tfrac12 x^2+3x+\tfrac{11}{2}.$

(c) Equation of $J$. From the roots $x=-5$ and $x=1$: $J:\;y=a(x+5)(x-1)$. The vertex has $x=-2$; imposing $y(-2)=3$: $a(3)(-3)=-9a=3\Rightarrow a=-\tfrac13$. Hence $J:\;y=-\tfrac13(x+5)(x-1)=-\tfrac13 x^2-\tfrac43 x+\tfrac53.$

(a) Vertices and intercepts. $R$: vertex $(-3;1)$; since it opens upward with $y_V=1>0$, it does not meet the $x$-axis; it meets the $y$-axis at $\left(0;\tfrac{11}{2}\right)$. $J$: vertex $(-2;3)$; meets the $x$-axis at $(-5;0)$ and $(1;0)$; meets the $y$-axis at $\left(0;\tfrac53\right)$.

(d) Difference. $R-J=\left(\tfrac12+\tfrac13\right)x^2+\left(3+\tfrac43\right)x+\left(\tfrac{11}{2}-\tfrac53\right)=\tfrac{5}{6}x^2+\tfrac{13}{3}x+\tfrac{23}{6}.$

(e) $\tfrac56 x^2+\tfrac{13}{3}x+\tfrac{23}{6}=0\;\Rightarrow\;5x^2+26x+23=0\;\Rightarrow\;x=\dfrac{-26\pm\sqrt{676-460}}{10}=\dfrac{-13\pm3\sqrt6}{5}\approx-1.13\ \text{and}\ -4.07.$ Setting $R-J=0$ is the same as imposing $R=J$: the solutions are therefore the abscissas of the intersection points of the two parabolas.

$\boxed{R:\;y=\tfrac12(x+3)^2+1;\quad J:\;y=-\tfrac13(x+5)(x-1);\quad R\cap J:\ x=\tfrac{-13\pm3\sqrt6}{5}}$