How do find the vertex and axis of symmetry for a quadratic equation #y=-2x^2-8x+3#?

1 Answer
Meave60
Jun 15, 2015

See the explanation below.

Explanation:

General equation for the quadratic formula: #y=ax^2+bx+c#

The graph of a quadratic equation is a parabola.The axis of symmetry is the vertical line that separates the parabola into two equal halves.The point on the x-axis where the vertical axis is placed is determined by the formula #x=(-b)/(2a)#

#y=ax^2+bx+c#

#y=-2x^2-8x+3#

#a=-2# and #b=-8#

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

#x=-2#

Substitute #-2# for #x# into the equation to find #y#.

#y=-2(-2)^2-8(-2)+3#

#y=-2(4)+16+3#

#y=-8+16+3=11#

#y=11#

The vertex of a parabola is the point where the parabola crosses its axis of symmetry. Vertex#=##(x,y)=(-2,11)#

http://hotmath.com/hotmath_help/topics/vertex-of-a-parabola.html
http://www.mathwarehouse.com/geometry/parabola/axis-of-symmetry.php

graph{y=-2x^2-8x+3 [-16.42, 15.6, -3.2, 12.82]}

Related questions
See all questions in Quadratic Functions and Their Graphs
Impact of this question
6086 views around the world
You can reuse this answer
Creative Commons License