How do you graph #y=-3x^2-4x#?

1 Answer
Jul 28, 2015

You'll get a parabola.

Explanation:

This function is a Quadratic (maximum degree of #x# is#=2#) so it will give you a PARABOLA.
Observing the coefficient of #x^2# you see that is #-3# which is #<0# so yours will be a downward parabola.
Now let us try to find interesting points of your parabola that will help us to plot it:

1] y-intercept:
set #x=0#; you get: #y=0#

2] x-intercept:
set #y=0#; you get #-3x^2-4x=0# solving it you find:
#-x(3x+4)=0#
so you have two solutions:
#x_1=0#
#x_2=-4/3#
so the x- intercepts will be:
#x=0, y=0#
#x=-4/3, y=0#

3] Vertex: this point represents the highest point reached by your parabola. the function can be written as #y=ax^2+bx+c#
where in your case:
#a=-3#
#b=-4#
#c=0#
The #x# coordinate of the vertex can then be found considering that #color(red)(x_v=-b/(2a))=4/(-6)=-4/6#;
The #y# coordinate of the vertex can then be found considering that #color(red)(y_v=-(Delta)/(4a))=-(b^2-4ac)/(4a)=(-16)/(-12)=4/3#

Graphically:
graph{-3x^2-4x [-11.25, 11.25, -5.625, 5.625]}