Question 128196
{{{f(x)=(x-6)^2 + 2}}} Start with the given function



To find the x-intercepts, simply let {{{f(x)=0}}}



{{{0=(x-6)^2 + 2}}} Plug in {{{f(x)=0}}}



{{{-2=(x-6)^2}}} Subtract 2 from both sides



{{{sqrt(-2)=x-6}}} Take the square root of both sides. 



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




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



{{{f(x)=(x-6)^2 + 2}}} Start with the given function



{{{f(0)=(0-6)^2 + 2}}} Plug in {{{x=0}}}



{{{f(0)=(-6)^2 + 2}}} Subtract



{{{f(0)=36 + 2}}} Square -6 to get 36



{{{f(0)=38}}} Add



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




Now because {{{f(x)=(x-6)^2 + 2}}} is in vertex form {{{f(x)=a(x-h)^2 + 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( 500, 500, -10, 10, -10, 40, (x-6)^2 + 2) }}} Graph of {{{f(x)=(x-6)^2 + 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.