calculus
posted by stephanie .
Let f(x)= Ax^2 + Bx + C. If f(x) passes through the point (1,1) and has a relative minimum at (1,13), find A, B, and C.

.) Find the tangent line to when . Find the y values on that tangent line when and then when and finally when . Compare these y values to the (approximate) function values at the same x values.

no you can't find the y calues by using the equation to the tangent line because of the missing values for a, b and c. the first derivative is 2Ax+B so obviously that doesnt work out. but when we set it equal to zero, which we are supposed to do when we know all the values, we can determine where it is increasing and decrasing. since we know the minimum is where the x=1, i can plug in one to get 2A+B=0 or 2=B/A however this still doesnt get me far. Also, looking at the equation, i know it has to be a parabola, and the question states it has a minimum i know that it is concave up everywhere. When i find the second derivative it is 2A threfore just a number, and proving that it is always concave up, but also states that A must be a positive number. and i really cant understand your answer at all... i think youre speaking jibberish. But, thaks for the attempt. I appreciate you trying to help.

just use the vertex method...
f(x) = a*(xb)+h
with :
a : what we're searching for (initially)
f(x) : 1 (from point (1;1))
x : 1 (from point (1;1))
b : x coordinate of the vertex : 1
h : y coordinate of the vertex : 13
=> 1 = a*(11)² 13
=> a = 10/4
fill in a in the ax²+bx+c=y equation, and replace x&y in one equation with (1;1) and the other equation with (1;13)
2 equations with 2 unknowns :
1) 14/4 x² + bx + c = y
2) 14/4 x² + bx + c = y
1) 14/4 (1)² + b*(1) + c = 1
2) 14/4 1² + b*1 + c = 13
1) 14/4 b +c = 1
2) 14/4 +b +c = 13
1) c = b(10/4) ==>
==>
2) (14/4) + b + b  (10/4) = 13
==>
2) 2b = 14
==>
b = 7
==>
1) c = (7)(10/4) = 19/2
a = 14/4
b = 7
c = 19/2
greetings from .be
==> final