What is the equation for the line of symmetry for the graph of the function #y=-4x^2+6x-8#?

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Answer 1 · 2016-06-07T09:25:27

The axis of symmetry is the line #x = 3/4#

The standard form for the equation of a parabola is

#y = ax^2 + bx + c#

The line of symmetry for a parabola is a vertical line. It can be found by using the formula #x = (-b)/(2a)#

In #y = -4x^2 + 6x -8, " "a = -4, b= 6 and c = -8#
Substitute b and c to get:

#x = (-6)/(2(-4)) = (-6)/(-8) = 3/4#

The axis of symmetry is the line #x = 3/4#

Answer 2 · 2016-06-07T10:27:18

#x = 3/4#

A parabola such as

#y = a_2x^2+a_1x+a_0#

can be put in the so called line of symmetry form by
choosing #c,x_0, y_0# such that

#y = a_2x^2+a_1x+a_0 equiv c(x-x_0)^2+y_0#

where #x = x_0# is the line of symmetry. Comparing coefficients we have

#{ (a_0 - c x_0^2 - y_0 = 0), (a_1 + 2 c x_0 = 0), (a_2 - c = 0) :}#

solving for #c, x_0, y_0#

# { (c = a_2), (x_0 = -a_1/(2 a_2)),( y_0 = (-a_1^2 + 4 a_0 a_2)/(4 a_2)) :} #

In the present case we have #c = -4, x_0 = 3/4, y_0 =-23/4# then

#x = 3/4# is the symmetry line and in symmetry form we have

#y = -4(x-3/4)^2-23/4#