How do you graph the parabola #y= 2x^2 – 4x – 6# using vertex, intercepts and additional points?

#y=2x^2-4x-6# is a quadratic equation in standard form:

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

where:

#a=2#, #b=-4#, #c=-6#

Vertex: maximum or minimum point of a parabola

Since #a>0#, the vertex is the minimum point and the parabola opens upward.

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

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

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

#x=4/4#

#x=1#

To find the y-coordinate of the vertex , substitute #1# for #x# in the equation and solve for #y#.

#y=2(1)^2-4(1)-6#

#y=2-4-6#

#y=-8#

The vertex is #(1,-8)#. Plot this point.

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

Substitute #0# for #x# and solve for #y#.

#y=2(0)^2-4(0)-6#

#y=-6#

The y-intercept is #(0,-6)#. Plot this point.

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

Substitute #0# for #y# and solve for #x#.

#2x^2-4x-6=0# #larr# I switched sides to get the variable on the left-hand side.

Factor out the common factor #2#.

#2(x^2-2x-3)=0#

Find two numbers that when multiplied equal #-3# and when added equal #-2#. The numbers #1# and #-3# meet the requirement.

#2(x+1)(x-3)=0#

Solve each binomial for #0#.

#x+1=0#

#x=-1#

#x-3=0#

#x=3#

The x-intercepts are #(-1,0)# and #(3,0)#. Plot the points.

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

Additional point 1

#x=2#

#y=2(2)^2-4(2)-6#

#y=8-8-6#

#y=-6#

Additional point 1 is #(2,-6)#. Plot this point.

Additional point 2

#x=-2#

#y=2(-2)^2-4(-2)-6#

#y=8+8-6#

#y=10#

Additional point 2 is #(-2,10)#. Plot this point.

Additional point 3

#x=4#

#y=2(4)^2-4(4)-6#

#y=32-16-6#

#y=10#

Additional point 3 is #(4,10)#. Plot this point.

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

graph{y=2x^2-4x-6 [-12.21, 10.29, -8.505, 2.745]}