Expert: Paul Klarreich - 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
Questioner: elle santy
Country: Australia
Category: Calculus
Private: No
Subject: calculus
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.
....................................................
Hi, Elle,

Let's do some translating. (into American English, unfortunately -- you may have to translate further)

The curve y=ax^2 + bx+c passes through the point (1,6).
......................>>
x = 1, y = 6  satisfies the equation, so:  6 = a + b + c
<<......................

The tangent there is parallel to the x axis,
......................>>
"there" means  at x = 1.
"The tangent [line] is parallel to the x axis"  means the slope is zero.
So the derivative [which is  2ax + b] at x = 1 is zero:
0 = 2a + b.
<<......................
whilst
...................
whilst???  Oh, yes -- Australian for 'while'.  Got it.
..................
the tangent at x= 3/2 on the curve is inclined at 45 degrees to the x axis.
..................>>
the derivative at x = 3/2 has slope equal to 1.  [I hope it is 1 -- you didn't say up or down, so I will assume up.]
2a(3/2) + b = 1,  or  3a + b = 1
<<................

Find the values of a, b, c
................>>
Solve these three simultaneous equations:
6 = a + b + c
0 = 2a + b
1 = 3a + b

I will leave that to you.
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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.

General procedure:

-- Find dy/dx.
-- Substitute  x = 2, to get the slope, m.
-- Find  y  at  x = 2.  [Gives the point  (x0,y0) ]
-- Use the point-slope form:  y - y0 = m(x - x0)

       (5-x)(4) - (4x + 7)(-1)
dy/dx = -------------------------  << quotient rule.
              (5-x)^2

       20 - 4x + 4x + 7
dy/dx = ------------------
           (5-x)^2

          27
dy/dx = --------
       (5-x)^2

At  x = 2:

         27        27
dy/dx = -------- = ---- = 3
       (5-2)^2      9

Also at  x = 2:

   4(2) + 7    15
y = --------- = --- = 5
    5 - 2       3

Now  use  x0 = 2,  y0 = 5,   m = 3.  

You can handle that.