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

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.

3 answers

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.
Aha! Thank you.
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.
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