Equation of a parabola from its elements and a parabola with a minimum

Text

From the figure a parabola is concave down, passes through the origin and through the point $(-6;0)$, with vertex $V\left(-3;\tfrac92\right)$; the line through the points $(0;1)$ and $(-3;3)$ is also drawn. (a) find the equation of the line; (b) find the vertex and the axis intercepts of the parabola; (c) find the equation of the parabola; (d) invent the equation of a second parabola that crosses the $x$-axis at the same points but has a minimum instead of a maximum.

Solution

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

(b) 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: $x=0\Rightarrow y=0$, the origin $(0;0)$.

(c) Equation of the parabola. Roots $x=0$ and $x=-6$, so $y=a\,x(x+6).$ The vertex has abscissa $x_V=-3$; imposing $y_V=\tfrac92$: $\frac92=a(-3)(-3+6)=-9a \;\Rightarrow\; a=-\frac12.$ Concave down ($a<0$), as required: $y=-\frac12x(x+6)=-\frac12x^2-3x.$

(d) Second parabola with a minimum. Same roots $x=0$ and $x=-6$ but concave up: just choose $a>0$. With $a=1$: $y=x(x+6)=x^2+6x,$ whose vertex (minimum) is at $\left(-3;-9\right)$. $\boxed{y=-\tfrac23x+1;\quad y=-\tfrac12x^2-3x;\quad V\left(-3;\tfrac92\right);\quad y=x^2+6x}$