Question 369244
{{{f(x)=7(x+1)^2+4}}}  I simplified it by FOILing the (x+1)^2 and multiplying 7 thru giving me

{{{f(x)=7x^2+14x+11}}}  The x-intercept occurs when f(x)=0.  
{{{0=7x^2+14x+11}}} Using the determinate {{{sqrt(b^2-4ac)}}} plugging in 7,14,11 for a,b,c.  you determine that There is no x intercept because you can't take the square root of a negative number.

The y intercept is found when x=0. So plug in 0 for any x.

{{{7(0)^2+14(0)+11=11}}} SO  the y intercept is at (0,11)

To find the vertex take the derivative of f(x) to get
{{{f'(x)=14x+14}}} set equal to 0 and solve for x.
{{{0=14x+14}}}
{{{x=-1}}}  Solve f(-1)

{{{f(-1)=7-14+11=4}}} So your vertex is at (-1,4) which is illustrated in the following graph
{{{graph(300,200,-5,5,-1,15,7(x+1)^2+4)}}}