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If the line and curve intersect, it happens when
[tex]mx + 2 = x^2-5x+18[/tex]
or
[tex]x^2-(m+5)x + 16 = 0[/tex]
Recall the discriminant (denoted by ∆) of a quadratic expression:
[tex]\Delta (ax^2+bx+c) = b^2 - 4ac[/tex]
If the discriminant is positive, then the quadratic has two real roots. If it's zero, it has only one real root. If it's negative, it has two complex roots. We're interested in the third case, because that would make it so the above equation has no real roots corresponding to points of intersection in the x,y-plane.
The discriminant here is
[tex](-(m+5))^2 - 4\cdot16 = (m+5)^2-64[/tex]
Find all m such that this quantity is negative:
[tex](m+5)^2-64 < 0 \\\\ \implies (m+5)^2 < 64 \\\\ \implies \sqrt{(m+5)^2} < \sqrt{64} \\\\ \implies |m+5| < 8 \\\\ \implies -8 < m + 5 < 8 \\\\ \implies \boxed{-13 < m < 3}[/tex]