Two parabolas: equations, intersections and difference parabola

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From the figure, two parabolas $B$ and $R$: $B$ is concave up, crosses the $x$-axis at $x=-4$ and $x=2$ and passes through $\left(0;-\tfrac{16}{9}\right)$; $R$ is concave down with vertex $V_R(-3;0)$ and passes through $(-1;-2)$. (a) find the vertices and axis intercepts of $B$ and $R$; (b) find the equation of $B$; (c) find the equation of $R$; (d) find the intersections between $B$ and $R$; (e) find the equation of the parabola $B-R$, its vertex and its axis intercepts.

Solution

(b) Equation of $B$. Roots $x=-4$ and $x=2$, so $y=a(x+4)(x-2)$. Passing through $\left(0;-\tfrac{16}{9}\right)$: $a(4)(-2)=-8a=-\frac{16}{9}\;\Rightarrow\;a=\frac29.$ $B:\; y=\frac29(x+4)(x-2)=\frac29\left(x^2+2x-8\right).$

(c) Equation of $R$. Vertex $(-3;0)$, so $y=a(x+3)^2$. Passing through $(-1;-2)$: $a(-1+3)^2=a\cdot4=-2\;\Rightarrow\;a=-\frac12.$ $R:\; y=-\frac12(x+3)^2.$

(a) Vertices and axis intercepts. $B$: vertex at $x=\tfrac{-4+2}{2}=-1$, $\;y=\tfrac29(1-2-8)=\tfrac29(-9)=-2$, so $V_B(-1;-2)$; $x$-axis intercepts $(-4;0)$ and $(2;0)$; $y$-axis intercept $\left(0;-\tfrac{16}{9}\right)$. $R$: vertex $V_R(-3;0)$; it meets the $x$-axis only at $x=-3$ (double root), $(-3;0)$; $y$-axis intercept $x=0\Rightarrow y=-\tfrac12(9)=-\tfrac92$, i.e. $\left(0;-\tfrac92\right)$.

(d) Intersections $B\cap R$. $\frac29\left(x^2+2x-8\right)=-\frac12(x+3)^2.$ Multiplying by $18$: $4\left(x^2+2x-8\right)=-9(x+3)^2$ $4x^2+8x-32=-9x^2-54x-81$ $13x^2+62x+49=0.$ $x=\frac{-62\pm\sqrt{62^2-4\cdot13\cdot49}}{2\cdot13}=\frac{-62\pm\sqrt{3844-2548}}{26}=\frac{-62\pm\sqrt{1296}}{26}=\frac{-62\pm36}{26}.$ So $x=-1$ (with $y=-2$) and $x=-\tfrac{98}{26}=-\tfrac{49}{13}\approx-3.769$.

(e) Difference parabola $B-R$. $B-R=\frac29\left(x^2+2x-8\right)+\frac12(x+3)^2.$ Expanding: $\tfrac29x^2+\tfrac49x-\tfrac{16}{9}+\tfrac12x^2+3x+\tfrac92$. Coefficients: $x^2:\;\frac29+\frac12=\frac{13}{18},\qquad x:\;\frac49+3=\frac{31}{9},\qquad \text{const.}:\;-\frac{16}{9}+\frac92=\frac{49}{18}.$ $B-R:\; y=\frac{13}{18}x^2+\frac{31}{9}x+\frac{49}{18}.$ Vertex: $x_V=-\dfrac{31/9}{2\cdot 13/18}=-\dfrac{31/9}{13/9}=-\dfrac{31}{13}\approx-2.385$. The discriminant is $\left(\frac{31}{9}\right)^2-4\cdot\frac{13}{18}\cdot\frac{49}{18}=\frac{961}{81}-\frac{2548}{324}=\frac{3844-2548}{324}=\frac{1296}{324}=4,$ so $y_V=-\dfrac{\Delta}{4a}=-\dfrac{4}{4\cdot 13/18}=-\dfrac{18}{13}\approx-1.385$. $x$-axis intercepts: $\tfrac{13}{18}x^2+\tfrac{31}{9}x+\tfrac{49}{18}=0\Rightarrow 13x^2+62x+49=0$, the same as in (d): $x=-1$ and $x=-\tfrac{49}{13}$. $y$-axis intercept: $x=0\Rightarrow y=\tfrac{49}{18}$. $\boxed{B:\,y=\tfrac29\left(x^2+2x-8\right);\ \ R:\,y=-\tfrac12(x+3)^2;\ \ B\cap R:\,x=-1,\,x=-\tfrac{49}{13};\ \ B-R:\,y=\tfrac{13}{18}x^2+\tfrac{31}{9}x+\tfrac{49}{18},\ V\left(-\tfrac{31}{13};-\tfrac{18}{13}\right)}$