How do you graph #y= 1( x-3)^2 -3#?

Question

Kennedy Adams · Accepted Answer

Showing how to determine the critical points. Needed for graph sketching. For a more precise plot will need to build a table of additional points.

#color(blue)("Important preamble")#

This is the vertex form equation of a quadratic in that

#y=ax^2+bx+c# is the same thing as #y=a(x+b/(2a))^2+c+k#

Where #k# is a correction for the introduced error consequential to 'forcing' #y=ax^2+bx+c# into this format.

Note that #a(b/(2a))^2+k=0#

The advantage of this vertex type equation is that with a little 'tweaking' you can virtually read of the coordinates for the vertex. Hence the name of 'vertex' form equation.

#x_("vertex")=(-1)xxb/(2a)#

#y_("vertex")=c+k#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Given that #y=1(x-3)^2-3# we have the following:

#color(blue)("Determining the critical points - the general shape")#

#color(blue)("As "a->1" is positive the general shape of the graph is "uu)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the critical points - the vertex")#

Given the equation:

#y=a(x+b/(2a))^2+c+k" "->" "y=1(xcolor(magenta)(-3))^2color(green)(-3)#

#x_("vertex")=(-1)xx(color(magenta)(-3))=+3#

#y_("vertex")=color(green)(-3)#

#=>color(blue)(" Verertex "->(x,y)=(+3,-3))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the critical points - the y-intercept")#

Expanding the backets:

#y=(x-3)^2-3" "->" "y=x^2-6x+9-3 #

#" "->" "color(green)(y=x^2-6xcolor(red)(+6))#

#" "y_("intercept")=color(red)(+6)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determining the critical points - the x-intercept")#

The x-axis is at #y=0#. So the graph crosses the x-axis at #y=0#

#y=(x-3)^2-3" "->" "0=(x-3)^2-3#

Add 3 to both sides

#3=(x-3)^2#

Take the square root of both sides

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

Add 3 to both sides

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

#x~~4.732 and x~~1.268 # to 3 decimal places
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If you need any more points you will need to build a table.

Nathan Axel · Answer

undefined To graph the equation $y = (x - 3)^2 - 3$, you need to follow these steps:

1. Identify the vertex: The vertex of the parabola is at the point $(3, -3)$.
2. Determine the direction of the parabola: Since the coefficient of $x$ is positive, the parabola opens upwards.
3. Find additional points: Choose additional points on either side of the vertex to plot on the graph.
4. Plot the points: Use the coordinates of the vertex and the additional points to plot on the coordinate plane.
5. Draw the parabola: Connect the points smoothly to form the parabolic curve.

The graph will be a parabola opening upwards with the vertex at $(3, -3)$.