Question

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

Answer 1

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

Explanation:

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

`x = 3/4`

Explanation:

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`