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:"

puu.sh)