What is the range of the graph of y = (x - 4)^2 + 3

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jimthompson5910 (jim_thompson5910):

Hint: In general, the range of y = a(x-h)^2+k is [k, infinity) if a > 0

OpenStudy (anonymous):

Y ≤ 3??

jimthompson5910 (jim_thompson5910):

close, it should be \[\Large y \ge 3\]

OpenStudy (anonymous):

What about this....
4x^2 + 13x = -3

jimthompson5910 (jim_thompson5910):

You need to complete the square. Do you know how do to that?

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OpenStudy (anonymous):

no

OpenStudy (anonymous):

So the answer would be x = -1/4 and x = -3

jimthompson5910 (jim_thompson5910):

you're looking for the range correct?

OpenStudy (anonymous):

No just solving the equation

jimthompson5910 (jim_thompson5910):

oh sry, I thought you were finding the range

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jimthompson5910 (jim_thompson5910):

4x^2 + 13x = -3
4x^2 + 13x + 3 = 0
Now use the quadratic formula to solve for x
x = (-b+-sqrt(b^2-4ac))/(2a)
x = (-(13)+-sqrt((13)^2-4(4)(3)))/(2(4))
x = (-13+-sqrt(169-(48)))/(8)
x = (-13+-sqrt(121))/8
x = (-13+sqrt(121))/8 or x = (-13-sqrt(121))/8
x = (-13+11)/8 or x = (-13-11)/8
x = -2/8 or x = -24/8
x = -1/4 or x = -3
So you are correct, nice job