How do you draw the graph of #y=3x^2-7#?

Note that #y = 3x^2 -7# is a parabola (which opens upward) so we will need to calculate a few sample points and then draw a parabolic curve through those points.

For example you might calculate #f(x)# for all integer values in the range #[--4,+4]# (note symmetry about the Y-axis reduces the number of actual calculations).
#(-4,41)# and #(+4,41)#
#(-3,20)# and #(+3,20)#
#(-2, 5)# and #(+2,5)#
#(-1,-4)# and #(+1,-4)#
#(0,-7)#

Mark these point on some graph paper and joint them in a smooth curve that should look something like:
graph{3x^2-7 [-9.71, 10.29, -7.16, 2.84]}

Given:

#y=3x^2-7# is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=3#, #b=0#, #c=-7#

To graph a parabola you need to determine the following points:

axis of symmetry, vertex, y-intercept (if applicable), x-intercepts (if applicable), and additional points if needed.

Axis of symmetry: vertical line that divides the parabola into two equal halves

Vertex: the maximum or minimum point of the parabola.

The formula for determining the axis of symmetry, which is also the #x#-coordinate of the vertex is:

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

#x=0/(2*3)#

#x=0#

To determine the #y#-coordinate by substituting #0# for #x# and solving for #y#.

#y=3(0)^2-7#

#y=-7#

The vertex is #(0,-7)#

Y-intercept: value of #y# when #x=0#.

In this particular case, the y-intercept is the same as the vertex, #(0,-7)#.

X-intercepts: values of #x# when #y=0#

#0=3x-7#

Add #7# to both sides.

#7=3x^2#

Divide both sides by #3#.

#7/3=x^2#

#+-sqrt(7/3)=x#

#x=-sqrt(7/3),# #sqrt(7/3)#

The x-intercepts are #(-sqrt(7/3),0)# and #(sqrt(7/3),0)#.

The approximate x-intercepts are #(-1.528,0)# and #(1.528,0)#.

Additional points: Choose values for #x# and solve for #y#.

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=3x^2-7 [-9.88, 10.12, -7.88, 2.12]}