SOLUTION: Can anyone help with the following problelm?
Determine the vertex of the parabola.
f(x) = x2 - 10x + 26
Algebra.Com
Question 441503: Can anyone help with the following problelm?
Determine the vertex of the parabola.
f(x) = x2 - 10x + 26
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
f(x) = x^2 - 10x + 26
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The vertex is on the line of symmetry
The line of symm is x = -b/2a
x = 10/2 = 5
f(5) = 25 - 50 + 26 = 1
Vertex at (5,1)
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Another method:
f(x) = x^2 - 10x + 26
f'(x) = 2x - 10 = 0
x = 5
f(5) .... as above
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Can anyone help with the following problelm?
Determine the vertex of the parabola.
f(x) = x^2 - 10x + 26
---
2 Methods: complete the square
and use a formula.
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Vertex occurs where x = -b/(2a) = 10/(2*1) = 5
f(x) = 5^2-10*5+26 = 1
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Vertex: (5,1)
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Cheers,
Stan H.
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