Find a parabola of the form given below that has slope m1 at x1, slope m2 at x2, and passes through the point P.

Question

y=ax^2 + bx + c
m1 = 8, x1 = 11
m2 = 7, x2 = 9
P = (5,7)

y = ?

I don't know where to begin. I can't find anywhere in my textbook that teaches how to solve this type of problem.

Answer 1

The slope of the parabola equals to the derivative of y:
y'(x)= dy/dx = d(ax^2 + bx + c)/dx
=2ax+b

Knowing that the slope at x1=11 is 8, we get
y'(11)=8
2a(11)+b=8 .....(1)
Similarly,
y'(9)=7
2a(9)+b=7.......(2)
From (1) and (2), solve for a and b.

c can be found by substituting the coordinates of x and y in P(5,7), i.e.
y(5)=7, or
a(5)²+b(5)+c = 7 ....(3)

Finally, from the values of a,b and c calculated above, substitute into the equations (1), (2) and (3) to make sure that all conditions are satisfied. If not, review the calculations.

Answer 2

Aha! Thank you.

Answer 3

Remember that the slope at a given point is equal to the first derivative at that point.

so if y = ax^2 + bx + c
dy/dx = 2ax + b

when x = 11, dy/dx = 8 ---> 22a + b = 8
when x = 9 , dy/dx = 7 ----> 18a + b = 7

subtract them ---> 4a = 1
a = 1/4
then 18(1/4) + b = 7
b = 5/2

so y = (1/4)x^2 + (5/2)x + c

plug in (5,7)

7 = (1/4)(25) + (5/2)(5) + c
c = -47/5

so finally
y = (1/4)x^2 + (5/2)x - 47/5

check my arithmetic.