Expert: Abe Mantell - 8/11/2009

Question
Please explain it so I understand

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

OK?

Abe