AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
The curve y=ax^2 + bx+c passes through the point (1,6). The tangent there is parallel to the x axis, whilst the tangent at x= 3/2 on the curve is inclined at 45 degrees to the x axis. Find the values of A, B, C
and................
Find the derivative of y= (4x+7)/(5-x) in the simplest form. Hence determine the equation of the tangent to the curve at the point where x=2.
Answer Hello,
Since (1,6) satisfies y=ax^2+bx+c, we get: 6=a+b+c...since the tangent line at x=1 is
horizontal, that means the turning point is at x=1 (and y=6), so we can write the parabola
as y=a(x-1)^2 + 6. At x=3/2 the tangent has a slope of 1, so y'(3/2)=1. y'=2a(x-1), so that
1=2a(3/2-1) ==> a=1...so we get y=(x-1)^2+6 or y=x^2-2x+7
If y=(4x+7)/(5-x), then y'=[4(5-x)-(4x+7)(-1)]/(5-x)^2=27/(5-x)^2
at x=2, y=15/3=5 and y'=27/9=3...tangent line at (2,5) with slope=3 is y=3x-1
