SOLUTION: Sketch the graph of f(x) = (x-6)^2 + 2. Label all x-intercept, y-intercepts and the vertex.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Sketch the graph of f(x) = (x-6)^2 + 2. Label all x-intercept, y-intercepts and the vertex.      Log On


   

Question 128196: Sketch the graph of f(x) = (x-6)^2 + 2. Label all x-intercept, y-intercepts
and the vertex.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=%28x-6%29%5E2+%2B+2 Start with the given function


To find the x-intercepts, simply let f%28x%29=0


0=%28x-6%29%5E2+%2B+2 Plug in f%28x%29=0


-2=%28x-6%29%5E2 Subtract 2 from both sides


sqrt%28-2%29=x-6 Take the square root of both sides.


Since you cannot take the square root of a negative number, there are no real x-intercepts. So this means that the graph does not cross the x-axis. So there are no x-intercepts.



Now to find the y-intercepts, simply plug in x=0


f%28x%29=%28x-6%29%5E2+%2B+2 Start with the given function


f%280%29=%280-6%29%5E2+%2B+2 Plug in x=0


f%280%29=%28-6%29%5E2+%2B+2 Subtract


f%280%29=36+%2B+2 Square -6 to get 36


f%280%29=38 Add


So when x=0, y=38. So the y-intercept is (0,38)



Now because f%28x%29=%28x-6%29%5E2+%2B+2 is in vertex form f%28x%29=a%28x-h%29%5E2+%2B+k where (h,k) is the vertex, we can see that the vertex is (6,2)



So putting all of this together, we get


+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+40%2C+%28x-6%29%5E2+%2B+2%29+ Graph of f%28x%29=%28x-6%29%5E2+%2B+2


and we can see that there are no x-intercepts, the graph has a y-intercept (0,38), and the vertex is (6,2). So this visually verifies our answer.