How do you graph the parabola # y = 2(x - 1)^2 - 5# using vertex, intercepts and additional points?

The vertex form of a quadratic is given as:

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

Where:

#bba# is the coefficient of #x^2#

#bbh# is the axis of symmetry.

#bbk# is the maximum/minimum value of the function.

The vertex of a parabola is the point where the curve turns, this is always at a maximum or a minimum value.

The coordinates of the vertex are:

#(h,k)#

Our function is in vertex form and from this we note that:

#h=1# and #k=-5#

So the coordinates of the vertex are:

#color(blue)((1,-5)#

Be careful when finding #h#. Remember that we have in the vertex form #(x-h)^2#, this means if we have #(x+2)^2#, then #h<0#, i.e. #h=-2#. #(x-(-2))^2=(x+2)^2#.

Other useful points for graphing are:

#y# axis intercept. This occurs where #x=0#:

#y=2((0)-1)^2-5#

#y=3#

Coordinates:

#color(blue)((0,3)#

#x# axis intercepts occur where #y=0#

#2(x-1)^2-5=0#

#(x-1)^2=5/2#

Taking roots:

#x-1=+-sqrt(5/2)#

#x=1+sqrt(5/2)=1+sqrt(10)/2#

#x=1-sqrt(5/2)=1-sqrt(10)/2#

Coordinates:

#color(blue)((1+sqrt(10)/2,0)# , #color(blue)((1-sqrt(10)/2,0)#

We know have four point in which to graph the function:

GRAPH: