.
#y=-3x^2+x-4#
This is a quadratic function and represents a parabola. Since the coefficient of #x^2# is negative the parabola opens down. This function is given in the form of:
#y=ax^2+bx+c#
From this form, we can find the #x#-coordinate of the vertex by using:
#x_("vertex")=(-b)/(2a)#
In our function:
#a=-3, b=1, c=-4#
#x_("vertex")=(-1)/(2(-3))=(-1)/(-6)=1/6#
This means the axis of symmetry which is a vertical line going through the vertex is:
#x=1/6#
The #y#-coordinate of the vertex can be calculated by plugging in its #x# into the function:
#y_("vertex")=-3(1/6)^2+1/6-4=-3(1/36)+1/6-4=-1/12+1/6-4=(-1+2-48)/12=-47/12#
Vertex #(1/6, -47/12)#
We can find the #x#-intercepts by setting the function equal to #0# and solving for #x#:
#-3x^2+x-4=0#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-1+-sqrt(1-4(-3)(-4)))/(2(-3))=1/6+-sqrt47/6i#
As evident, we get two imaginary values for #x#. This means there are no #x#-intercepts. And because the parabola opens down, it is entirely below the #x#-axis.
We set #x=0# to solve for the #y#-intercept.
#y=-3(0)+(0)-4#
#y=-4#,
The #y#-intercept is:
#(0, -4)#
The graph of this function is:
The parabola is in red and the axis of symmetry is in green.
