SOLUTION: Does this function have a maximum or a minimum? {{{x^2+4x-5=0}}} After solving this equation I came up with (x=1) and (x=-5). Am I correct? Please include details. I don't yet kno
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-> SOLUTION: Does this function have a maximum or a minimum? {{{x^2+4x-5=0}}} After solving this equation I came up with (x=1) and (x=-5). Am I correct? Please include details. I don't yet kno
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Question 178187: Does this function have a maximum or a minimum? After solving this equation I came up with (x=1) and (x=-5). Am I correct? Please include details. I don't yet know how to find a max or min. Please explain how to find this answer. Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Given:
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Simply by looking at the coefficient associated with the 'x^2' term (in this case, it's a POSITIVE 1) you can immediately tell that you will find the MINIMUM by finding the "vertex" or x-axis of symmetry.
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A quadratic forms a parabola -- either it is "open upward" or "open downward". If it is "open upward" the vertex is at the minimum. If is "open downward" the vertex is at a maximum. If the coefficient (associated with the x^2) is POSITIVE -- it is "open upward" (think of it this way, if you're POSITIVE you would have a happy face). If the coefficient is NEGATIVE -- it is "open downward" (if you're NEGATIVE, you would be sad faced).
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The "vertex form" is:
y= a(x-h)^2+k
where (h,k) is the vertex
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To find the vertex, complete the square:
(h,k) = (-2, -9)
