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)
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
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