What is the axis of symmetry and vertex for the graph #f(x)= 2x^2 - 11#?

Question

Isaiah Anthony · Accepted Answer

undefined The axis of symmetry for the graph of $ f(x) = 2x^2 - 11 $ is the vertical line represented by the equation $ x = \frac{-b}{2a} $, where $ a $ is the coefficient of the $ x^2 $ term (in this case, $ 2 $) and $ b $ is the coefficient of the $ x $ term (which is $ 0 $ since there's no $ x $ term). So, the axis of symmetry is $ x = \frac{0}{2(2)} = 0 $.

To find the vertex, plug the value of $ x $ from the axis of symmetry equation into the original function to find the corresponding $ y $ value. So, $ f(0) = 2(0)^2 - 11 = -11 $. Therefore, the vertex is at the point $ (0, -11) $.

Savannah Anderson · Answer

Vertex#->(x,y)=(0,-11)#

The axis of symmetry is the y-axis

First write as #" "y=2x^2+0x-11# Then write as #" "y=2(x^2+0/2x)-11# This is part of the process for completing the square. I have written this format on purpose so that we can apply: The value for #x_("vertex")= (-1/2)xx(+0/2)=0# So the axis of symmetry is the y-axis. So #y_("vertex")=2(x_("vertex"))^2-11# #y_("vertex")=2(0)^2-11# #y_("vertex")=-11# Vertex#->(x,y)=(0,-11)#

Julia Adams · Answer

Axis of symmetry is #y#-axis

Vertex is at # (0,-11)# From the equation given it is obvious that vertex is at # x=0 ,y=-11#. and the axis of symmetry is #x=0# that is the #y#- axis. There is no #x# term so the graph has not moved left or right, only down #11# units.

Abigail Allen · Answer

undefined The axis of symmetry for the graph of $ f(x) = 2x^2 - 11 $ is the vertical line that passes through the vertex. To find the axis of symmetry, you use the formula $ x = -\frac{b}{2a} $, where $ a $ is the coefficient of the $ x^2 $ term (in this case, $ 2 $) and $ b $ is the coefficient of the $ x $ term (which is $ 0 $ because there's no $ x $ term). Substituting these values into the formula, we get $ x = -\frac{0}{2(2)} = 0 $. Therefore, the axis of symmetry is $ x = 0 $.

To find the vertex, you substitute the value of the axis of symmetry into the original function to find the corresponding $ y $ value. So, $ f(0) = 2(0)^2 - 11 = -11 $. Therefore, the vertex is at the point $ (0, -11) $.