How do you find the vertex of $y = 4 x - x^{2}$ $y=4x-x^2$ ?

Question

How do you find the vertex of $y = 4 x - x^{2}$ $y=4x-x^2$ ?

1 Answer

Impact of this question

2912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-06T02:43:02

(2, 4)

Explanation:

The equation $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ can be modified to the standard form
${(x - α)}^{2} = \pm 4 p (y - β)$ $(x-alpha)^2=+-4p(y-beta)$ , gviing vertex $(α, β)$ $(alpha, beta)$ and parameter p for the size of the parabola.
This way, the vertex of the parabola $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ is $(- \frac{b}{2 a}, c - \frac{b^{2}}{4 a})$ $(-b/(2a), c-b^2/(4a))$ . .
Here, $a = - 1, b = 4 and c = 0$ $a=-1, b=4 and c=0$ .

Of course, direct manipulation for the standard form is easier.

Science

Math

Humanities

... and beyond

How do you find the vertex of $y = 4 x - x^{2}$ $y=4x-x^2$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the vertex of y=4x−x2y=4x-x^2?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the vertex of $y = 4 x - x^{2}$ $y=4x-x^2$ ?