What is the vertexof #y= -5x^2 − 3x #?

2 Answers
Mar 26, 2018

Vertex: #(frac{-3}{10},frac{9}{20})#

Explanation:

First, use the axis of symmetry formula #(AoS: x = frac{-b}{2a})# to find the x-coordinate of the vertex #(x_{v})# by substituting #-5# for #a# and #-3# for #b#:

#x_{v} = frac{-b}{2a}#

#x_{v} = frac{-(-3)}{2(-5)}#

#x_{v} = frac{-3}{10}#

Then find the y-coordinate of the vertex #(y_{v})# by substituting #frac{-3}{10}# for #x# in the original equation:

#y_{v} = -5x^{2}-3x#

#y_{v} = -5(frac{-3}{10})^{2}-3(frac{-3}{10})#

#y_{v} = -5(frac{9}{100})+frac{9}{10}#

#y_{v} = frac{-45}{100}+frac{90}{100}#

#y_{v} = frac{45}{100}#

#y_{v} = frac{9}{20}#

Finally, express the vertex as an ordered pair:

Vertex: #(x_{v},y_{v}) = (frac{-3}{10},frac{9}{20})#

Mar 26, 2018

The vertex is #(-3/10,9/20)# or #(-0.3,0.45)#.

Explanation:

Given:

#y=-5x^2-3x# is a quadratic equation in standard form:

#ax^2+bx-3x#,

where:

#a=-5#, #b=-3#, #c=0#

The vertex of a parabola is its maximum or minimum point. In this case, since #a<0#, the vertex will be the maximum point and the parabola will open downward.

To find the #x#-value of the vertex, use the formula for the axis of symmetry:

#x=(-b)/(2a)#

#x=(-(-3))/(2*(-5))#

#x=3/(-10)#

#x=-3/10#

To find the #y#-value of the vertex, substitute #-3/10# for #x# and solve for #y#.

#y=-5(-3/10)^2-3(-3/10)#

Simplify.

#y=-color(red)cancel(color(black)(5))^1(9/color(red)cancel(color(black)(100))^20)+9/10#

#y=-9/20+9/10#

Multiply #9/10# by #2/2# to get the common denominator #20#.

#y=-9/20+9/10xx2/2#

#y=-9/20+18/20#

#y=9/20#

The vertex is #(-3/10,9/20)# or #(-0.3,0.45)#.

graph{y=-5x^2-3x [-10, 10, -5, 5]}