What is the vertex form of #y=5x^2+4x+7#?

1 Answer
Shwetank Mauria
Feb 27, 2016

#y=5(x+2/5)^2+31/5#, where vertex is #(-2/5,31/5)#

Explanation:

Vertex form of equation is of type #y = a(x – h)^2 + k#, where #(h,k)# is the vertex. For this, in the equation #y=5x^2+4x+7#, one should first take #5# out out of first two terms and then make it complete square, as follows:

#y=5x^2+4x+7=5(x^2+4/5x)+7#

To make #(x^2+4/5x)#, complete square, one has to add and subtract, 'square of half the coefficient of #x#, and thus this becomes

#y=5x^2+4x+7=5(x^2+4/5x+(2/5)^2)+7-5*(2/5)^2# or

#y=5(x+2/5)^2+7-4/5# or

#y=5(x-(-2/5))^2+31/5#, where vertex is #(-2/5,31/5)#

Related questions
See all questions in Vertex Form of a Quadratic Equation
Impact of this question
1993 views around the world
You can reuse this answer
Creative Commons License