1. ## quadratic function problems :'(

Q: The quadratic function which takes the value 41 at x=-2 and the value 20 at x=5 and is minimized at x=2 is

y = __x2 - __x + __

The minimum value of this function is __

I really need to solve this . thank you

2. ## Re: quadratic function problems :'(

Let y = ax^2 + bx + c for some unknown a, b and c. Write the given conditions as equations in a, b and c. You'll get three equations in three variables. Recall that the point of minimum of a parabola is -b / (2a).

3. ## Re: quadratic function problems :'(

that will be the equation but i don't get how to compute it. I'm sorry I'm totally clueless

4. ## Re: quadratic function problems :'(

Originally Posted by pauliana
that will be the equation
What will be the equation?

Originally Posted by pauliana
but i don't get how to compute it.
You don't compute an equation; you solve it.

Start by substituting x = -2 into ax^2 + bx + c = 41. Similarly, substitute x = 5 into ax^2 + bx + c = 20. Finally, write the equation saying that the point of minimum, i.e., x = -b / (2a), equals 2.

5. ## Re: quadratic function problems :'(

a(-2)^2 + b(-2) + c = 41 = -4a - 2b + c = 41

a(5)^2 + b(5) + c = 20 = 25a + 5b + c = 20

is this correct?

6. ## Re: quadratic function problems :'(

Correct except that (-2)^2 = 4 and not -4. Negative times negative gives positive. Now you need the third equation, and then you can solve the resulting system of equations.

7. ## Re: quadratic function problems :'(

it this right?

4a - 2b + c = 41
25a + 5b + c = 20
____________________
29a - 3b + 2c = 61

8. ## Re: quadratic function problems :'(

Originally Posted by pauliana
it this right?

4a - 2b + c = 41
25a + 5b + c = 20
____________________
29a - 3b + 2c = 61
No, 5b - 2b = 3b, not -3b. Also, you don't gain anything by adding the equations because the resulting equation still has all three variables. It would be better to subtract the equations; then c would be eliminated.

Originally Posted by emakarov
Now you need the third equation, and then you can solve the resulting system of equations.

9. ## Re: quadratic function problems :'(

4a - 2b + c = 41
- 25a + 5b + c = 20
_____________________
-21a + 3b =21

right?
wait how about The minimum value of this function is ___ ?

10. ## Re: quadratic function problems :'(

Originally Posted by pauliana
4a - 2b + c = 41
- 25a + 5b + c = 20
_____________________
-21a + 3b =21
Hmm, no. The minus in the second equation looks like it acts on 25a only. In any case, -2b - 5b = -7b, so the result should be -21a -7b = 21. You can divide both sides by 7.

Originally Posted by pauliana
wait how about The minimum value of this function is ___ ?

Originally Posted by emakarov
Recall that the point of minimum of a parabola is -b / (2a).
Originally Posted by emakarov
Finally, write the equation saying that the point of minimum, i.e., x = -b / (2a), equals 2.

11. ## Re: quadratic function problems :'(

7 [-21a -7b = 21] 7 will be -3a = 7

then

x = -b
______ = 2
2a

I'm so sorry I'm lost again

12. ## Re: quadratic function problems :'(

I found the answer sheet for this question and it says

y = 3x^2 - 12x + 5

then the value of D = -7

I don't get it

13. ## Re: quadratic function problems :'(

First, it is better to avoid notation spanning several lines because this forum does not preserve alignment. Instead of $x=\frac{-b}{2a}=2$, you can simply write x = -b / (2a) = 2.

I don't understand the notation "7 [-21a -7b = 21] 7". Dividing -21a - 7b = 21 by 7 gives -3a - b = 3. The last equation, -b / (2a) = 2, can be multiplied by 2a to get -b = 4a, or 4a + b = 0. Thus, you have

\begin{align*}-3a - b &= 3\\4a + b &= 0\end{align*}

Adding these equations gives a = 3. Substituting this value of a into either of the two equations above gives you b, and knowing a and b you can find c from any of the equations in post #5 (with the correction in post #6).

14. ## Re: quadratic function problems :'(

Originally Posted by pauliana
I found the answer sheet for this question and it says

y = 3x^2 - 12x + 5

then the value of D = -7

I don't get it
The minimum value D of this quadratic function is the value at the point of minimum (see the end of post #10).

15. ## Re: quadratic function problems :'(

oh I see I get it now . I do apologize for all the repeated question it just that I totally forget how to solve quadratic functions and the parabola itself

thank you very very much

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