How do you graph the parabola #y = -2(x -1)^2 + 3# using vertex, intercepts and additional points?

1 Answer
Aug 3, 2018

Refer to the explanation.

Explanation:

Graph:

#y=-2(x-1)^2+3#

This is a quadratic equation in vertex form:

#y=a(x-h)^2+k#,

where:

#a=-2#, #h=1#, #k=3#

The vertex is the point #(h,k)#, which is #(1,3)#. Plot this point.

The y-intercept is the value of #y# when #x=0#. Substitute #0# for #x# and solve for #y#.

#y=-2(0-1)^2+3#

#y=-2(-1)^2+3#

#y=-2(1)+3#

#y=-2+3#

#y=1#

The y-intercept is #(0,1)#. Plot this point.

To find the x-intercepts, convert the equation from vertex form to standard form.

Expand #(x-1)^2# to #x^2-2x+1#.

#y=-2(x^2-2x+1)+3#

#y=-2x^2+4x-2+3#

#y=-2x^2+4x+1#

The x-intercepts are the values of #x# when #y=0#. Substitute #0# for #y# and solve for #x#.

#0=-2x^2+4x+1#,

where:

#a=-2#, #b=4#, #c=1#

Switch sides.

#-2x^2+4x+1=0#

Solve for the x-intercepts using the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Plug in the known values.

#x=(-4+-sqrt(4^2-4*-2*1))/(2*-2)#

Simplify.

#x=(-4+-sqrt(24))/(-4)#

Prime factorize #24#.

#x=(-4+-sqrt(2xx2xx2xx3))/(-4)#

Simplify #2xx2# to #2^2#.

#x=(-4+-sqrt(2^2xx2xx3))/(-4)#

Apply rule #sqrt(a^2)=a#

#x=(-4+-2sqrt6)/(-4)#

#x=(-4+2sqrt6)/(-4)#, #(-4-2sqrt6)/(-4)#

Simplify.

#x=1-(2sqrt6)/2#, #1+(2sqrt6)/2#

The x-intercepts are #(1-(sqrt6)/2,0)# and #(1+(sqrt6)/2,0)#.

The approximate x-intercepts are #(-0.225,0)# and #(2.225,0)#.

Plot the x-intercepts.

Additional point

Choose values for #x# and solve for #y#.

#x=2#

#y=-2(2-1)^2+3#

#y=-2(1)^2+3#

#y=-2(1)+3#

#y=-2+3#

#y=1#

The additional point is #(2,1)#. Plot this point.

To find additional points, choose values for #x# and solve for #y#.

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

graph{y=-2(x-1)^2+3 [-10, 10, -5, 5]}