Question 514633: I am an adult college student that has returned to college after 26 years. I am currently taking the second part of a college algebra class. I have done pretty well up until this point and this problem. I am stuck and have consulted several friends, references, and sites, but cannot grasp how to solve this problem. I am a visual learner, so I need to see every step of the process in order to understand the concept. Would somebody be able to assist me with this problem and how to solve it.
Thank you.
Very confused and frustrated.
Problem #1:
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function.
The x-coordinates of vertex is?
This is what I did so far.
f(x)= -2x^2 + 2x + 3
= -2(x^2 -x) + 3
= -2 (x^2 - x + 1/4 - 1/4) + 3
= -2 (x^2 - x + 1/4) + -2(-1/4) + 3
= -2 (x^2 -x +1/4) + and this is where I get totally lost and confused.
Thank you very much for your assistance.
Sincerely,
Christine Fitzgerald
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website!
Find the vertex, the line of symmetry, the maximum or minimum value of the quadratic function, and graph the function.
The x-coordinates of vertex is?
This is what I did so far.
f(x)= -2x^2 + 2x + 3
= -2(x^2 -x) + 3
= -2 (x^2 - x + 1/4 - 1/4) + 3
= -2 (x^2 - x + 1/4) + -2(-1/4) + 3
= -2 (x^2 -x +1/4) + and this is where I get totally lost and confused.
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= -2 (x^2 - x + 1/4) + -2(-1/4) + 3
You were almost done
= -2(x - 1/2)^2 + 7/2
The line of symmetry is x = 1/2
The vertex is (1/2,7/2), a maximum
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A better method is:
The LOS is x = -b/2a = -2/-4
x = 1/2
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The vertex is (1/2,f(1/2)) = (1/2,7/2)
And you're done.
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To graph this, and others, dl the FREE graph software at
http://www.padowan.dk
Use Insert, then enter the function
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