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Prove that the parabola

y = x^{2} + 10 x + 2

lies entirely above the line

y = 4 x - 8

Hint

Is the parabola convex or concave?

Hint

Consider the discriminant of

x^{2} + 10 x + 2 = 4 x - 8

Solution

FConsider the discriminant of

\begin{aligned} x^{2} + 10 x + 2 & = 4 x - 8 \\ x^{2} + 6 x + 10 & = 0 \end{aligned}

which is

\begin{aligned} Δ & = 6^{2} - 4 (1) (10) \\ = - 4 \\ < 0 \end{aligned}

The discriminant is negative, so the parabola and line never intersect.

Now let's let $x = 0$ . For the parabola, $y = 2$ and for the line $y = - 8$ . Therefore the point on the parabola lies above the point on the line. But since the parabola and line never intersect, then all points on the parabola must lie above the line.