Parabola as a product of two lines

Text

Given the lines Agnese $\left(y=-\dfrac{x}{2}+2\right)$ and Lucia $\left(y=\dfrac{x}{3}-1\right)$, and the parabola Gertrude obtained by setting $y$ equal to the product of the right-hand sides: $y=\left(-\dfrac{x}{2}+2\right)\left(\dfrac{x}{3}-1\right)$.

• (a) draw Agnese and Lucia, find their axis intersections and the vertex of Gertrude;
• (b) find the intersection of Agnese and Lucia;
• (c) find the intersections of Gertrude with each line. What do you notice?
• (d) find a parabola with the same $x$-intercepts as Gertrude passing through the intersection of Agnese and Lucia.

Solution

Expand Gertrude: $y=-\dfrac{x^2}{6}+\dfrac{7x}{6}-2$. (a) Agnese $\cap x$: $(4;0)$, $\cap y$: $(0;2)$. Lucia $\cap x$: $(3;0)$, $\cap y$: $(0;-1)$. Vertex of Gertrude: $x_V=\dfrac72$, $y_V=\dfrac14\cdot\dfrac16=\dfrac{1}{24}\approx0{,}042$, so $V\!\left(\dfrac72;\dfrac{1}{24}\right)$. (b) $-\dfrac x2+2=\dfrac x3-1\Rightarrow x=\dfrac{18}{5}$, $y=\dfrac15$: point $\left(\dfrac{18}{5};\dfrac15\right)$. (c) Gertrude$\cap$Agnese: $\left(-\dfrac x2+2\right)\left(\dfrac x3-2\right)=0\Rightarrow x=4$ or $x=6$: points $(4;0)$ and $(6;-1)$. Gertrude$\cap$Lucia: $x=3$ or $x=2$: points $(3;0)$ and $\left(2;-\dfrac13\right)$. Remark: Gertrude meets each line where that line crosses the $x$-axis (since $\text{Gertrude}=\text{Agnese}\cdot\text{Lucia}$ vanishes where a factor vanishes): the zeros of Gertrude are $x=4$ and $x=3$. (d) Parabola with zeros $x=4,3$: $y=k(x-4)(x-3)$ through $\left(\dfrac{18}{5};\dfrac15\right)$: $k=-\dfrac56$. $\boxed{y=-\frac56(x-4)(x-3);\qquad V_\text{Gertrude}\!\left(\tfrac72;\tfrac{1}{24}\right);\qquad \text{Agnese}\cap\text{Lucia}=\left(\tfrac{18}{5};\tfrac15\right).}$