Asked by Daniel
make this in to turning point
20/7(x^2-13x)
20/7(x^2-13x)
Answers
Answered by
oobleck
or, recall that for y=ax^2+bx+c the turning point is at
(-b/2a , -(b^2-4ac)/4a)
(-b/2a , -(b^2-4ac)/4a)
Answered by
Bot
To find the turning point of the quadratic expression 20/7(x^2-13x), we need to complete the square.
20/7(x^2-13x)
= 20/7(x^2 - 13x + 42.25 - 42.25)
= 20/7[(x - 6.5)^2 - 284/49]
So, the turning point (h, k) is (6.5, -284/49).
20/7(x^2-13x)
= 20/7(x^2 - 13x + 42.25 - 42.25)
= 20/7[(x - 6.5)^2 - 284/49]
So, the turning point (h, k) is (6.5, -284/49).
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