Parabola and line, intersections and relative position

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Given the parabola Gina $\left(y=-\dfrac{x^2}{4}+2x+5\right)$ and the line Lina $\left(y=2x-4\right)$:

• (a) draw the graphs and find their intersections with the axes and the vertex of Gina;
• (b) find graphically the intersection points of Gina and Lina;
• (c) find them with algebra;
• (d) find for which $x$ the points of Gina lie above those of Lina.

Solution

(a) Gina. Vertex: $x_V=4$, $y_V=9$, i.e. $V(4;9)$. $x$-axis: $x^2-8x-20=0\Rightarrow x=10,\,-2$; $y$-axis: $(0;5)$. Lina. $\cap x$: $(2;0)$; $\cap y$: $(0;-4)$. (c) Gina–Lina: $-\dfrac{x^2}{4}+2x+5=2x-4\Rightarrow x^2=36\Rightarrow x=\pm6$. For $x=6$: $(6;8)$; for $x=-6$: $(-6;-16)$. (d) Gina above Lina: $-\dfrac{x^2}{4}+9>0\Rightarrow -6<x<6$. $\boxed{V(4;9);\quad (6;8)\ \text{and}\ (-6;-16);\quad \text{Gina}>\text{Lina}\ \text{for}\ -6<x<6.}$