What is the vertex form of #y=-2x^2 + 2x+3#?

Question

Ethan Adkins · Accepted Answer

#y=(-2)(x-1/2)^2+3 1/2#

The general vertex form is: #color(white)("XXX")y=m(x-a)^2+b#

Given: #color(white)("XXX")y=-2x^2+2x+3#

Extract the #m# component: #color(white)("XXX")y=(-2)(x^2-1x)+3#

Complete the square #color(white)("XXX")y=(-2)(x^2-1x[+(1/2)^2])+3[-(-2)(1/2)^2]#

#color(white)("XXX")y=(-2)(x-1/2)^2+ 3 1/2#

which is the vertex form with vertex at #(1/2, 3 1/2)#

graph{-2x^2+2x+3 [-1.615, 3.86, 1.433, 4.17]}

Eva Abramson · Answer

undefined The vertex form of a quadratic function $ y = ax^2 + bx + c $ is given by $ y = a(x - h)^2 + k $, where $ (h, k) $ represents the coordinates of the vertex.

To find the vertex form of the quadratic function $ y = -2x^2 + 2x + 3 $:

1. Factor out the leading coefficient from the quadratic term: $ y = -2(x^2 - x) + 3 $.
2. Complete the square inside the parentheses: $ y = -2(x^2 - x + \frac{1}{4} - \frac{1}{4}) + 3 $.
3. Rewrite the expression: $ y = -2\left[(x - \frac{1}{2})^2 - \frac{1}{4}\right] + 3 $.
4. Distribute and simplify: $ y = -2(x - \frac{1}{2})^2 + \frac{1}{2} + 3 $.
5. Combine constants: $ y = -2(x - \frac{1}{2})^2 + \frac{7}{2} $.

So, the vertex form of the function $ y = -2x^2 + 2x + 3 $ is $ y = -2(x - \frac{1}{2})^2 + \frac{7}{2} $.