Question 978065
The tangent to the curve y=ax^2+bx+2 at (1,0.5) is parallel to the normal to the
curve y=x^2+6x+4 at (-2,-4).Find the value of a and b.
<pre>
To prevent students from simply taking our answers here and submitting them for
a grade and learning nothing from them, here is a problem exactly in every
detail like the one you submitted but with different numbers and different
answers.  But every step is explained:
</pre>
The tangent to the curve y=ax^2+bx+36.5 at (6,-2.5) is parallel to the normal to
the curve y=x^2-4x-4 at (3,-7).Find the value of a and b.
<pre>
We begin by drawing the graph of y=x^2-4x-4, with a tangent (in green) and a
normal line (in blue) at (3,-7): 

{{{drawing(325,400,-3,10,-10,6,
locate(3.3,-6.5,"(3,-7)"),
graph(325,400,-3,10,-10,6,x^2-4x-4), green(line(-13,-37.3,4.5,-4)),
blue(line(-7,-2,15,-13))
 )}}}

We need the slope of the green tangent line so that we can find the slope of the
blue normal line by taking its reciprocal with the opposite sign.

The derivative IS an expression for the slope of the tangent line at any point,
so we find the derivative of:

 y = x^2-4x-4
y' = 2x-4

and evaluate it at the point (3,-7), using only its x value x=3

y'<sub>@x=3</sub> = 2(3)-4 = 6-4 = 2, the slope of the green tangent line.

That means that the slope of the blue normal line is its reciprocal with the
opposite sign, or -1/2.

Now we know that the graph of y = ax^2+bx+36.5 goes through the point (6,-2.5),
since it is tangent there.  So we draw a line parallel to the blue line through
(6,-2.5) which will be tangent to the graph of y = ax^2+bx+36.5 at (6,-2.5).

{{{drawing(325,400,-3,10,-10,6,
locate(3.3,-6.5,"(3,-7)"),
graph(325,400,-3,10,-10,6,x^2-4x-4),
locate(4.5,-3,"(6,-2.5)"),

green(line(-13,-39,4.5,-4)),



circle(6,-2.5,0.15),circle(6,-2.5,0.13),circle(6,-2.5,0.11),circle(6,-2.5,0.09),circle(6,-2.5,0.07),circle(6,-2.5,0.05),circle(6,-2.5,0.03),circle(6,-2.5,0.01),




blue(line(-7,-2,15,-13),line(-9,5,11,-5))
 )}}}

And we can approximately sketch in the graph of y = ax^2+bx+36.5 so that it
will be tangent to the upper blue line at (6,-2.5):

{{{drawing(325,400,-3,10,-10,6,
locate(3.3,-6.5,"(3,-7)"),
graph(325,400,-3,10,-10,6,x^2-4x-4,x^2-(25/2)x+73/2),
locate(4.5,-3,"(6,-2.5)"),

circle(6,-2.5,0.15),circle(6,-2.5,0.13),circle(6,-2.5,0.11),circle(6,-2.5,0.09),circle(6,-2.5,0.07),circle(6,-2.5,0.05),circle(6,-2.5,0.03),circle(6,-2.5,0.01),



green(line(-13,-39,4.5,-4)),

blue(line(-7,-2,15,-13),line(-9,5,11,-5))
 )}}}

Now as in the other graph, we know that the derivative IS the slope of the
tangent line, so we find the derivative of

 y = ax^2+bx+36.5
y' = 2ax+b

and evaluate it at the point (6,-2.5), using only its x value x=6
y'<sub>@x=6</sub> = 2a(6)+b = 12a+b, the slope of the tangent line to
y = ax^2+b+36.5 at (6,-2.5) which is the slope of the upper blue tangent line.

So we set that slope 12a+b equal to -1/2

12a+b = -1/2, multiplying through by 2:

eq. (1)       24a+2b = -1

We also know that y = ax^2+b+36.5 goes through the point (6,-2.5) so we
substitute x=6 and y=-2.5 into that equation

   y = ax^2+bx+36.5
-2.5 = a(6)^2+b(6)+36.5
-2.5 = a(36)+6b+36.5
-2.5 = 36a+6b+36.5
 -39 = 36a+6b

Divide through by 3 and reverse left and right sides

eq. (2)     12a+2b=-13

Now we solve the system of equations eq. (1) and eq. (2)

{{{system(24a+2b = -1,12a+2b=-13)}}}

We multiply the second equation through by -1 and add to eliminate b:

{{{system(24a+2b = -1,-12a-2b=13)}}}

12a = 12
  a = 1

Substituting in eq. (1)

24a+2b = -1
24(1)+2b = -1
  24+2b = -1
      2b = -25
       b = -25/2

Answer:  a = 1,  b = -25/2 = -12.5

[So the equation of the green graph is y = x^2-12.5x+36.5]

Now do your problem, which is exactly in every detail like this one.

Edwin</pre>