Derivative?

Question

A parabola in the form y=ax^2+bx+c passes through the point (2,5). The line y=-8x+29 is also tangent to the parabola at the point (4,-3). Find the equation of the parabola algebraicly.

I have tried but can't do it please help.

Answer 1

the slope of the prabola and the tangent will be the same at the point of contact
y=ax^2+bx+c
y'=2ax+b
at 4,-3
-8=2a(4)+b
=>8a+b=-8
parabola passes through (2,5)
a(2)^2+b(2)+c=5
4a+2b+c=5
passes through (4,-3)
a(4)^2+b(4)+c=-3
16a+4b+c=-3
4a+2b+c=5
12a+2b=-8
8a+b=-8
16a+2b=-16
12a+2b=-8
subtracting
4a=-8
a=-2
substituting
-16+b=-8
b=8
substituting
4a+2b+c=5
-8+16+c=5
c=-3
equation of the parabola is
=-2x^2+8x-3

Answer 2

because
y = -8x + 29 at (4,-3)
c = 29

for the point (4,-3):
-3 = 16a + 4b + 29
-3 = 4(4a + b) + 29
-32 = 4(4a +b)
-8 = 4a + b (equation Q)

for the point (2,5):
5 = 4a + 2b
5 - 29 = 4a + 2b
-24 = 4a + 2b
-12 = 2a + b
=> b = -2a - 12

substituting for b in equation Q:
4a -2a -12 = -8
2a - 12 = -8
2a = 4

which implies:
a = 2
b = -16
c = 29

OR:

y = 2x^2 -16x + 29

comment to raj:
you say -8=2a(4)+b
where does the -8 come from?