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

Jun 7, 2016

The axis of symmetry is the line $x = \frac{3}{4}$

Explanation:

The standard form for the equation of a parabola is

$y = a {x}^{2} + b x + c$

The line of symmetry for a parabola is a vertical line. It can be found by using the formula $x = \frac{- b}{2 a}$

In $y = - 4 {x}^{2} + 6 x - 8 , \text{ } a = - 4 , b = 6 \mathmr{and} c = - 8$
Substitute b and c to get:

$x = \frac{- 6}{2 \left(- 4\right)} = \frac{- 6}{- 8} = \frac{3}{4}$

The axis of symmetry is the line $x = \frac{3}{4}$

Jun 7, 2016

$x = \frac{3}{4}$

Explanation:

A parabola such as

$y = {a}_{2} {x}^{2} + {a}_{1} x + {a}_{0}$

can be put in the so called line of symmetry form by
choosing $c , {x}_{0} , {y}_{0}$ such that

$y = {a}_{2} {x}^{2} + {a}_{1} x + {a}_{0} \equiv c {\left(x - {x}_{0}\right)}^{2} + {y}_{0}$

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 $c = - 4 , {x}_{0} = \frac{3}{4} , {y}_{0} = - \frac{23}{4}$ then

$x = \frac{3}{4}$ is the symmetry line and in symmetry form we have

$y = - 4 {\left(x - \frac{3}{4}\right)}^{2} - \frac{23}{4}$