Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.