How do you find the vertex and intercepts for #y=1/2(x-8)^2+2#?

1 Answer
Mar 22, 2018

#Vertex=(8, 2)#
#y"-intercept:" (0, 34)#
#x"-intercept: None"#

Explanation:

Quadratic equations are either shown as:

#f(x)=ax^2+bx+c# #color(blue)(" Standard Form")#

#f(x)=a(x-h)^2+k# #color(blue)(" Vertex Form")#

In this case, we'll ignore the #"standard form"# due to our equation being in #"vertex form"#

#"Vertex form"# of quadratics is much easier to graph due to there not being a need to solve for the vertex, it's given to us.

#y=1/2(x-8)^2+2#

#1/2= "Horizontal stretch"#
#8=x"-coordinate of vertex"#
#2=y"-coordinate of vertex"#

It's important to remember that the vertex in the equation is #(-h, k)# so since h is negative by default, our #-8# in the equation actually becomes positive. That being said:

#Vertex = color(red)((8, 2)#

Intercepts are also very easy to calculate:

#y"-intercept:"#

#y=1/2(0-8)^2+2# #color(blue)(" Set " x=0 " in the equation and solve")#

#y=1/2(-8)^2+2# #color(blue)(" "0-8=-8)#

#y=1/2(64)+2# #color(blue)(" "(-8)^2=64)#

#y=32+2# #color(blue)(" "1/2*64/1=64/2=32)#

#y=34# #color(blue)(" "32+2=4)#

#y"-intercept:"# #color(red)((0, 34)#

#x"-intercept:"#

#0=1/2(x-8)^2+2# #color(blue)(" Set " y=0 " in the equation and solve")#

#-2=1/2(x-8)^2# #color(blue)(" Subtract 2 from both sides")#

#-4=(x-8)^2# #color(blue)(" Divide both sides by " 1/2)#

#sqrt(-4)=sqrt((x-8)^2)# #color(blue)(" Square-rooting both removes the square")#

#x"-intercept:"# #color(red)("No Solution")# #color(blue)(" Can't square root negative numbers")#

You can see this to be true, as there are no #x"-intercepts:"#

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