What is the vertex form of y=-x^2+5xy=−x2+5x?

2 Answers
kimchiboy03
Dec 21, 2017

(x - 5/2)^2 - 25/4(x−52)2−254

Explanation:

To find the vertex form, you need to complete the square:

-x^2 + 5x−x2+5x
= x^2 - 5x=x2−5x
= x^2 - 5x + (5/2)^2 - (5/2)^2=x2−5x+(52)2−(52)2
= (x - 5/2)^2 - (5/2)^2=(x−52)2−(52)2
= (x - 5/2)^2 - 25/4=(x−52)2−254

Nallasivam V
Dec 21, 2017

y=-(x-5/2)^2+25/4y=−(x−52)2+254

Explanation:

Given -

y=-x^2+5xy=−x2+5x

Vertex

x=(-b)/(2a)=(-5)/(-1xx2)=5/2x=−b2a=−5−1×2=52

At x=5/2x=52;

y=-(5/2)^2+5(5/2)=-25/4+25/2=(-25+50)/4=25/4y=−(52)2+5(52)=−254+252=−25+504=254

Vertex (5/2, 25/4)(52,254)

The vertex form of the quadratic equation is -

y=a(x-h)^2+ky=a(x−h)2+k

Where -

a=-1a=−1 - coefficient of x^2x2
h=5/2h=52 - x - coordinate of the vertex
k=25/4k=254 - y - coordinate of the vertex

Substitute these values in the formula

y=-1(x-5/2)^2+25/4y=−1(x−52)2+254

y=-(x-5/2)^2+25/4y=−(x−52)2+254

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