How do you use the important points to sketch the graph of #y= ½(x-4)(x+2)#?

Question

Eva Abramson · Accepted Answer

x-intercepts #->(x,y)=(-2,0) and (4,0)#

Vertex#->(x,y)=(1,-9/2)#

#y_("intercept")->(x,y)=(0,-4)#

#color(blue)("Determine the general shape of the graph")#

If you multiply out the brackets you have #y=+1/2x^2+....#
As the coefficient of #x^2# is positive we have the general shape of #uu#. Thus the vertex is a minimum.

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#color(blue)("Determine the x-intercepts")#
The x-intercept is at #y=0#

This indicates that we have

#(x-4)=0=>x=+4#
#(x+2)=0=>x=-2#
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#color(blue)("Determine the vertex")#

#x_("vertex")# will be mid point between the x-intercepts so we have:

#x_("vertex")=(-2+4)/2=1#

Substitute for #x# to obtain #y_("vertex")#

#y_("vertex")=1/2(1-4)(1+2)=-9/2#

#y_("vertex")=-9/2#

Vertex#->(x,y)=(1,-9/2)#
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#color(blue)("Determine y-intercept")#

Consider the standard form #y=ax^2+bx+c#

The y-intercept is at #x=0# giving:

#y_("intercept")=c#

So from #y=1/2(x-4)(x+2) # we have:

# c=1/2xx(-4)xx2=1/2xx(-8) = -4#

#y_("intercept")->(x,y)=(0,-4)#

Eva Adamson · Answer

undefined To sketch the graph of $ y = \frac{1}{2}(x - 4)(x + 2) $, follow these steps:

1. Identify the important points: the x-intercepts, the y-intercept, and any vertex or turning points.
2. Find the x-intercepts by setting $ y = 0 $ and solving for $ x $.
3. Find the y-intercept by setting $ x = 0 $ and solving for $ y $.
4. Determine the vertex or turning point using the formula $ \frac{-b}{2a} $ for quadratic equations in the form $ ax^2 + bx + c $.
5. Plot these points on a graph.
6. Use the shape of the parabola to sketch the curve, ensuring it passes through the identified points.