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*(x-b)+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*(-1-1)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
To find the values of A, B, and C, we can use the information given in the problem.
First, we know that the function passes through the point (-1,1). This means that when x=-1, the value of f(x) is 1. Plugging these values into the equation of the function, we get:
1 = A(-1)^2 + B(-1) + C
1 = A - B + C (Equation 1)
Next, we are given that the function has a relative minimum at (1,-13). This means that when x=1, the value of f(x) is -13. Plugging these values into the equation of the function, we get:
-13 = A(1)^2 + B(1) + C
-13 = A + B + C (Equation 2)
We now have a system of two equations with three unknowns (A, B, and C). To solve this system, we can use any method, such as substitution or elimination. In this case, we will use the elimination method.
Adding Equation 1 and Equation 2 together, we get:
1 + (-13) = (A - B) + (A + B) + (C + C)
-12 = 2A + 2C
Dividing both sides by 2, we have:
-6 = A + C (Equation 3)
Now, we can substitute Equation 3 into either Equation 1 or Equation 2 to determine the values of the remaining variables. Let's substitute it into Equation 1:
1 = A - B + C
Substituting -6 for A + C, we have:
1 = (-6) - B
1 = -6 - B
Adding B to both sides, we get:
B + 1 = -6
Subtracting 1 from both sides:
B = -7
Now, we can substitute the value of B into Equation 1 to find A:
1 = A - (-7) + C
1 = A + 7 + C
Subtracting 7 from both sides:
-6 = A + C
Comparing this equation to Equation 3, we can see that the values of A and C are the same. Therefore, we can conclude that:
A = -6
B = -7
C = -6
So, the values of A, B, and C are -6, -7, and -6, respectively.