SOLUTION: What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + 8t^2 + 55t + 41?
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Question 1209301: What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + 8t^2 + 55t + 41?
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
Answer: -23/9
This is approximately equal to -2.55556 where the 5's go on forever but we have to round at some point.
Explanation
I'll use x in place of t.
x^2 -9x - 36 + 8x^2 + 55x + 41
= (x^2 + 8x^2) + (-9x + 55x) + ( -36 + 41)
= 9x^2 + 46x + 5
The vertex of y = ax^2+bx+c is located at (h,k) where h = -b/(2a)
y = 9x^2 + 46x + 5 has the coefficients a = 9, b = 46, c = 5
h = -b/(2a)
h = -46/(2*9)
h = -46/18
h = (-2*23)/(2*9)
h = -23/9
This is the x coordinate of the vertex.
This x value will make 9x^2 + 46x + 5 as small as possible.
Note that a = 9 is positive, so the parabola opens upward, which places the vertex at the lowest point.
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Another approach would be to use the quadratic formula to solve 9x^2 + 46x + 5 = 0 to get the roots x = -5 and x = -1/9
I'll leave the scratch work for the student to do.
Or you could factor 9x^2 + 46x + 5 into (x+5)(9x+1) to be able to see the roots easily.
x = -5 and x = -1/9 are where the parabola crosses the x axis.
The midpoint of the x intercepts is the x coordinate of the vertex.
This is due to the parabola's mirror symmetry about the center line.
p,q are the roots
h = x coordinate of the vertex (h,k)
h = (1/2)*(p+q)
h = (1/2)*( -5 + (-1/9) )
h = (1/2)*( -45/9 + (-1/9) )
h = (1/2)*(-46/9)
h = (-46)/(2*9)
h = (-2*23)/(2*9)
h = -23/9
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