calculus

posted by .

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.

  • calculus -

    .) 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.

  • calculus -

    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.

  • calculus -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. algebra

    1. How to find the numbers, if there are any, at which f has a realtive minimum. What is the realive minima?
  2. calculus

    If g is a differentiable function such that g(x) is less than 0 for all real numbers x and if f'(x)=(x2-4)g(x), which of the following is true?
  3. Calculus AB

    Find a, b, c, and d such that the cubic function ax^3 + bx^2 + cx + d satisfies the given conditions Relative maximum: (2,4) Relative minimum: (4,2) Inflection point: (3,3) So this is what I have so far: f'(x) = 3a^2 + 2bx + c f''(x) …
  4. Calculus

    Sketch a graph of the parabola y=x^2+3. On the same graph, plot the point (0,−6). Note there are two tangent lines of y=x2+3 that pass through the point (0,−6). The tangent line of the parabola y=x^2+3 at the point (a,a^2+3) …
  5. calculus

    I needed help with these FRQ in my APCalc course. Any help or walkthrough would be extremely helpful - thanks in advance. Let f be the function given by f(x)=3ln((x^2)+2)-2x with the domain [-2,4]. (a) Find the coordinate of each relative …
  6. Calculus

    Let f(x) be a polynomial function such that f(-2)=5, f'(-2)=0, and f"(-2)=3. The point (-2,5) is a(n)____of the graph f. A. Relative maximum B. Relative minimum C. Intercept D. Point of Inflection E. None of these
  7. Calculus

    Let f(x) be a polynomial function such that f(-2)=5, f’(-2)=0 and f”(-2)=3. The point (-2, 5) is which of the following for the graph of f?
  8. Calculus

    Let f(x) be a polynomial function such that f(4)=-1, f’(4)=2 and f”(4)=0. If x<4, then f”(x)<0 and if x>4, then f”(x)>0. The point (-4, -1) is which of the following for the graph of f?
  9. Calculus

    Find the values of x that give relative extrema for the function f(x)=3x^5-5x^3 A. Relative maximum: x= 1; Relative minimum: x=sqrt(5/3) B. Relative maximum: x=-1; Relative minimum: x=1 C. Relative maxima: x=+or- 1; Relative minimum: …
  10. Calculus check

    Let f(x) be a polynomial function such that f(3)=3, f'(3)=0 and f"(3)=-3. What is the point (3,3) on the graph y=f(x)?

More Similar Questions