Determine a quadratic function f(x)=ax^2 + bx + c if its graph passes through the point (2,19) and has a horizontal tangent at the point (-1, 8)

Answer 1

first of all you got 2 points, so
19 = 4a + 2b + c
8 = a - b + c
subtract them:
11 = 3a + 3b

f ' (x) = 2ax + b
at x = -1 this is zero, so
2a(-1) + b = 0
-2a + b = 0 or b = 2a
plug into 3a + 3b = 11
3a + 6a = 11
a = 11/9
b = 22/9
go into one of the first equations to solve for c

was expecting nicer numbers, check my arithmetic

Answer 2

if the tangent is horizontal, the slope is zero
so, you want 2ax+b = 0
At x = -1 that means
2a = b
Also, since (-1,8) is on the curve, that means that
a - 2a + c = 8
-a+c = 8
Since (2,19) is on the curve,
4a + 4a + c = 19
9a+c = 19
so, a = 11/9
b = 22/9
c = 83/9
f(x) = 11/9 x^2 + 22/9 x + 83/9

Or, eschewing calculus, we know the vertex is at (-1,8) so
y = a(x+1)^2 + 8
Plugging in (2,19) we get
9a+8 = 19
a = 11/9
y = 11/9 (x+1)^2 + 8
which is the same as we got above