How do you write $f(x)=-2x^2+5x+9$ in vertex form?

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Answer 1 · 2016-08-25T15:07:49

$y = -2(x - 5/4)^2 + 97/8$ is the equation of the function in vertex form.

You must complete the square to achieve this task.

$y = -2(x^2 - 5/2x) + 9$

$y = -2(x^2 - 5/2x + n - n) + 9$

$n = (b/2)^2$

$n = ((-5/2)/2)^2$

$n = 25/16$

$y = -2(x^2 - 5/2x + 25/16 - 25/16) + 9$

$y = -2(x^2 - 5/2x + 25/16) + 25/8 + 9$

$y = -2(x - 5/4)^2 + 97/8$

We are now in the form $y = a(x - p)^2 + q$ . Mission accomplished!

Hopefully this helps!

How do you write f(x)=-2x^2+5x+9f(x)=-2x^2+5x+9 in vertex form?