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Help! I really need some help!

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For what constant k is 1 the minimum value of the quadratic 3x^2 - 15x + k over all real values of x? (x cannot be nonreal)

Guest Jul 2, 2017
edited by Guest  Jul 2, 2017

#1
+5931
+3

Let's say

y  =  3x2 - 15x + k                              Now let's get this into vertex form. Divide through by 3.

y/3  =  x2 - 5x + k/3                            Subtract  k/3  from both sides of this equation.

y/3 - k/3  =  x2 - 5x                             Add  (5/2)2  ,  or   25/4   to both sides of this equation.

y/3 - k/3 + 25/4  =  x2 - 5x + 25/4      Now we can factor the right side...

y/3 - k/3 + 25/4  =  (x - 5/2)2              Multiply through by  3  .

y - k + 75/4  =  3(x - 5/2)2                  Add  k  and subtract  75/4  to both sides.

y  =  3(x - 5/2)2 + k - 75/4

The vertex form of a parabola is:     y  =  a(x - h)2 + j     , where  (h, j)  is the vertex of the parabola.

(Normally you see k instead of j, but I used j since we already have a k in this problem.)

If the minimum value of  y  is  1, then the y-coordinate of the vertex of this parabola will = 1.

This means....

k - 74/4   =  1

k   =   1 + 75/4   =   79/4   =   19.75

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Here is another approach...

y  =  3x2 - 15x + k             Take the derivative with respect to x on both sides.

dy/dx  =  6x - 15               The x value when dy/dx = 0 will be the x coordinate of the minimum,

since this is a positive parabola. So, set it equal to 0 and solve for x.

0  =  6x - 15

x  =  15/6  =  5/2     Plug this value for x into the original equation to find the y coordinate of the min.

y  =  3(5/2)2 - 15(5/2) + k

y  =  -75/4 + k                   We are told that the y coordinate of the min. = 1,

so plug in  1 for y and solve for k.

1  = -75/4 + k

k  =  1 + 75/4  =  19.75

hectictar  Jul 2, 2017
edited by hectictar  Jul 2, 2017
Sort:

#1
+5931
+3

Let's say

y  =  3x2 - 15x + k                              Now let's get this into vertex form. Divide through by 3.

y/3  =  x2 - 5x + k/3                            Subtract  k/3  from both sides of this equation.

y/3 - k/3  =  x2 - 5x                             Add  (5/2)2  ,  or   25/4   to both sides of this equation.

y/3 - k/3 + 25/4  =  x2 - 5x + 25/4      Now we can factor the right side...

y/3 - k/3 + 25/4  =  (x - 5/2)2              Multiply through by  3  .

y - k + 75/4  =  3(x - 5/2)2                  Add  k  and subtract  75/4  to both sides.

y  =  3(x - 5/2)2 + k - 75/4

The vertex form of a parabola is:     y  =  a(x - h)2 + j     , where  (h, j)  is the vertex of the parabola.

(Normally you see k instead of j, but I used j since we already have a k in this problem.)

If the minimum value of  y  is  1, then the y-coordinate of the vertex of this parabola will = 1.

This means....

k - 74/4   =  1

k   =   1 + 75/4   =   79/4   =   19.75

_________________________________________________________________________________

_________________________________________________________________________________

Here is another approach...

y  =  3x2 - 15x + k             Take the derivative with respect to x on both sides.

dy/dx  =  6x - 15               The x value when dy/dx = 0 will be the x coordinate of the minimum,

since this is a positive parabola. So, set it equal to 0 and solve for x.

0  =  6x - 15

x  =  15/6  =  5/2     Plug this value for x into the original equation to find the y coordinate of the min.

y  =  3(5/2)2 - 15(5/2) + k

y  =  -75/4 + k                   We are told that the y coordinate of the min. = 1,

so plug in  1 for y and solve for k.

1  = -75/4 + k

k  =  1 + 75/4  =  19.75

hectictar  Jul 2, 2017
edited by hectictar  Jul 2, 2017
#2
+1

Thank you so much!

Guest Jul 2, 2017
#3
+81022
+1

Very nice, hectictar....I probably would favor the Calculus approach, but the other is more comprehensible for the  Algebra student.....

CPhill  Jul 2, 2017
#4
+5931
+3

Thank you CPhill !!      I'm glad you said that because I was thinking at first that it might have been unnecessary to put down the calculus method.

hectictar  Jul 2, 2017

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