Parabola and intersection with a line

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Given the parabola $p:y=\dfrac{1}{4}x^2-\dfrac{1}{2}x-1$ and the line $r:\dfrac{2}{5}x+\dfrac{1}{5}y+1=0$:

• (a) find the vertex of $p$;
• (b) the intercepts of $p$ with the axes;
• (c) sketch it;
• (d) find the intersections between $p$ and $r$;
• (e) decide whether $B(1;0)$ belongs to $p$.

Solution

Coefficients of $p$: $a=\dfrac14$, $b=-\dfrac12$, $c=-1$; $a>0$ opens upward.

(a) Vertex. $x_V=-\frac{b}{2a}=-\frac{-\frac12}{2\cdot\frac14}=1, \qquad y_V=\frac14-\frac12-1=-\frac54=-1.25$ so $V(1;-1.25)$.

(b) Intercepts with the axes. $x$-axis: $\dfrac14x^2-\dfrac12x-1=0 \Rightarrow x^2-2x-4=0 \Rightarrow x=1\pm\sqrt5\approx 3.236,\ -1.236$. $y$-axis: $x=0 \Rightarrow y=-1$, i.e. $(0;-1)$.

(c) Graph. Parabola opening upward, vertex $V(1;-1.25)$, crossing the $x$-axis at $1\pm\sqrt5$ and the $y$-axis at $(0;-1)$.

(d) Intersections $p\cap r$. The line $r$, multiplied by $5$, is $2x+y+5=0$, i.e. $y=-2x-5$. Substituting into $p$ and multiplying by $4$: $\frac14x^2-\frac12x-1=-2x-5 \;\Longrightarrow\; x^2-2x-4=-8x-20 \;\Longrightarrow\; x^2+6x+16=0$ The discriminant is $\Delta=36-64=-28<0$: no real solution, so $p$ and $r$ do not intersect.

(e) Does $B(1;0)$ belong? $y(1)=-1.25\neq0$, so $B$ does not belong to $p$. $\boxed{V(1;-1.25)\quad (0;-1)\quad p\cap r=\varnothing\ (\Delta<0)\quad B\notin p}$