Two parabolas: vertices, intercepts and distance between the vertices

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Given the parabolas $L:y=\dfrac{x^2}{3}+x-3$ and $G:y=-\dfrac{x^2}{4}+4$:

• (a) find the vertices;
• (b) the intercepts with the $x$-axis;
• (c) the intercepts with the $y$-axis;
• (d) sketch them;
• (e) find the distance between the vertices of $L$ and $G$.

Solution

(a) Vertices. For $L$: $a=\dfrac13$, $b=1$, so $x_{V_L}=-\dfrac{b}{2a}=-\dfrac{1}{2/3}=-\dfrac32$ and $y_{V_L}=\frac13\left(-\tfrac32\right)^2+\left(-\tfrac32\right)-3=\frac34-\frac32-3=-\frac{15}{4}=-3.75$ so $V_L\left(-\tfrac32;-\tfrac{15}{4}\right)$. For $G$: $a=-\dfrac14$, $b=0$, so $x_{V_G}=0$ and $y_{V_G}=4$, i.e. $V_G(0;4)$.

(b) Intercepts with the $x$-axis. $L$: $\dfrac{x^2}{3}+x-3=0 \Rightarrow x^2+3x-9=0 \Rightarrow x=\dfrac{-3\pm\sqrt{45}}{2}=\dfrac{-3\pm3\sqrt5}{2}\approx 1.854,\ -4.854$. $G$: $-\dfrac{x^2}{4}+4=0 \Rightarrow x^2=16 \Rightarrow x=\pm4$: points $(-4;0)$ and $(4;0)$.

(c) Intercepts with the $y$-axis (set $x=0$). $L$: $y=-3 \Rightarrow (0;-3)$. $G$: $y=4 \Rightarrow (0;4)$.

(d) Graph. $L$ opens upward, vertex $V_L(-1.5;-3.75)$; $G$ opens downward, vertex $V_G(0;4)$.

(e) Distance between the vertices. $d(V_L,V_G)=\sqrt{\left(0+\tfrac32\right)^2+\left(4+\tfrac{15}{4}\right)^2}=\sqrt{\tfrac94+\left(\tfrac{31}{4}\right)^2}=\sqrt{2.25+60.0625}=\sqrt{62.3125}\approx 7.894$ $\boxed{V_L\left(-\tfrac32;-\tfrac{15}{4}\right)\quad V_G(0;4)\quad d(V_L,V_G)\approx7.894}$