SOLUTION: Find the maximum/minimum value and the value of x when it occurs by completing the square. a) y=5x^2-40x+81 b) y=-3x^2-30x-79

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find the maximum/minimum value and the value of x when it occurs by completing the square. a) y=5x^2-40x+81 b) y=-3x^2-30x-79      Log On


   

Question 1151571: Find the maximum/minimum value and the value of x when it occurs by completing the square.
a) y=5x^2-40x+81
b) y=-3x^2-30x-79
Found 3 solutions by Boreal, MathLover1, MathTherapy:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
5x^2-40x+81
divide by 5
5(x^2-8x+81/5)
complete the square
5[(x-4)^2+16+(81/5-80/5)], 80/5 being 16, which has to be subtracted
5[(x-4)^2+(1)]
vertex is at (4, 1), and it is a minimum value of 80-160+81=1
graph%28300%2C300%2C-1%2C6%2C-5%2C5%2C5x%5E2-40x%2B81%29
-3x^2-30x-79
-3[(x^2+10x+25-(79/3))]
-3[(x+5)^2-(4/3)]
vertex is at (-5, -4)
graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-3x%5E2-30x-79%29

Answer by MathLover1(20849) About Me  (Show Source):
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a)
y=5x%5E2-40x%2B81
y=%285x%5E2-40x%29%2B81
y=5%28x%5E2-8x%29%2B81........complete square
y=5%28x%5E2-8x%2Bb%5E2%29-5b%5E2%2B81
y=5%28x%5E2-8x%2B4%5E2%29-5%2A4%5E2%2B81
y=5%28x-4%29%5E2-5%2A16%2B81
y=5%28x-4%29%5E2-80%2B81
y=5%28x-4%29%5E2%2B1
the value of x is 4

b)
y=-3x%5E2-30x-79
y=-%283x%5E2%2B30x%29-79
y=-3%28x%5E2%2B10x%2Bb%5E2%29-%28-3%29b%5E2-79
y=-3%28x%5E2%2B10x%2B5%5E2%29%2B3%2A5%5E2-79
y=-3%28x%2B5%29%5E2%2B3%2A25-79
y=-3%28x%2B5%29%5E2%2B75-79
y=-3%28x%2B5%29%5E2-4
the value of x is -5


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Find the maximum/minimum value and the value of x when it occurs by completing the square.
a) y=5x^2-40x+81
b) y=-3x^2-30x-79
When you have completed the square on a), you should get the VERTEX form of a quadratic equation: matrix%281%2C3%2C+y%2C+%22=%22%2C+5%28x+-+4%29%5E2+%2B+1%29, which you now need to compare to the 

equation: matrix%281%2C3%2C+y%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29, with (h, k) or in this case, (4, 1) being the vertex. 

And, with the given equation having an "a" value that's > 0, this parabola will have a highlight_green%28matrix%281%2C6%2C+MINIMUM%2C+VALUE%2C+of%2C+y%2C+%22=%22%2C+1%29%29, since k = 1. 

You should now be able to follow this concept and answer b)!