Finding the equation of a parabola and of a line

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A parabola and a line are drawn in the Cartesian plane. From the figure one reads that the parabola has vertex $V\left(-3;-\tfrac92\right)$ and passes through the origin and through the point $(-6;0)$, while the line passes through the points $(0;1)$ and $(3;-1)$. (a) find the equation of the line; (b) find the equation of the parabola; (c) find the coordinates of the vertex and the intercepts of the parabola with the axes; (d) find the intersections between the line and the parabola.

Solution

(a) Equation of the line. The line passes through $(0;1)$ and $(3;-1)$. Its slope is $m=\frac{-1-1}{3-0}=\frac{-2}{3}=-\frac23.$ Since the intercept is $q=1$ (it passes through $(0;1)$), $y=-\frac23x+1.$

(b) Equation of the parabola. The parabola passes through the roots $x=0$ and $x=-6$, so it has the form $y=a\,x(x+6).$ The vertex has abscissa $x_V=-3$; imposing $y_V=-\tfrac92$: $-\frac92=a(-3)(-3+6)=a(-3)(3)=-9a \;\Rightarrow\; a=\frac12.$ Hence $y=\frac12x(x+6)=\frac12x^2+3x.$

(c) Vertex and axis intercepts. Vertex: $V\left(-3;-\tfrac92\right)$. $x$-axis intercepts: $x=0$ and $x=-6$, i.e. $(0;0)$ and $(-6;0)$. $y$-axis intercept: $x=0\Rightarrow y=0$, i.e. the origin $(0;0)$.

(d) Line–parabola intersections. $\frac12x^2+3x=-\frac23x+1.$ Multiplying by $6$: $3x^2+18x=-4x+6 \;\Rightarrow\; 3x^2+22x-6=0.$ $x=\frac{-22\pm\sqrt{22^2+4\cdot3\cdot6}}{2\cdot3}=\frac{-22\pm\sqrt{484+72}}{6}=\frac{-22\pm\sqrt{556}}{6}.$ Since $\sqrt{556}=2\sqrt{139}$, this simplifies to $x=\frac{-11\pm\sqrt{139}}{3}.$ Numerically $x\approx0.263$ and $x\approx-7.596$; the ordinates follow by substituting into the line $y=-\tfrac23x+1$. $\boxed{y=-\tfrac23x+1;\quad y=\tfrac12x^2+3x;\quad V\left(-3;-\tfrac92\right);\quad x=\frac{-11\pm\sqrt{139}}{3}}$