What is the axis of symmetry and vertex for the graph #y=4x^2-2x+2#?

Question

Genesis Allen · Accepted Answer

Axis of symmetry: #x=1/4#

Vertex is at #(1/4, 1 3/4)# The equation of a parabola is #y = ax^2 +bx+c# #y =4x^2 - 2x+2# is a the equation of a parabola To find the axis of symmetry use: #x = (-b)/(2a)# #x = (-(-2))/(2(4)) = 2/8 = 1/4# Therefore, the #x#-co-ordinate of the vertex is #1/4#. Substitute #1/4# into the equation to find the #y#-value. #y = 4(1/4)^2-2(1/4)+2# #y = 4xx1/16 -2/4+2# #y = 1/4-2/4+2# #y = 1 3/4# Vertex is #(1/4, 1 3/4)#

Eva Adamson · Answer

undefined The axis of symmetry for the graph $ y = 4x^2 - 2x + 2 $ is the vertical line passing through the vertex. The formula to find the axis of symmetry is $ x = \frac{-b}{2a} $, where $ a $ is the coefficient of the $ x^2 $ term, and $ b $ is the coefficient of the $ x $ term. For this equation, $ a = 4 $ and $ b = -2 $, so the axis of symmetry is $ x = \frac{-(-2)}{2 \cdot 4} = \frac{1}{4} $.

To find the vertex, substitute the value of $ x $ obtained for the axis of symmetry into the equation of the parabola to find the corresponding $ y $ value. So, $ y = 4\left(\frac{1}{4}\right)^2 - 2\left(\frac{1}{4}\right) + 2 = 2 $. Therefore, the vertex of the parabola is $ (1/4, 2) $.