h(x) = x4 - 2x3 - 6x2 + 8x + 5 

y = h(-2) = (-2)4 - 2(-2)3 - 6(-2)2 + 8(2) + 5
y = h(-2) = 16 - 2(8) - 6(4) + 16 + 5
y = h(-2) = 16 - 16 - 24 + 21
y = h(-2) = 0 - 24 + 21
y = h(-2) = -3  

Do that also with x = -3,-1, 0, 1, 2, 3, 4
and make this table:

 x |y=h(x)     point
---------------------
-3 | 62       (-3,62)   <-- too high to plot
-2 | -3       (-2,-3)
-1 | -6       (-1,-6)
 0 |  5        (0,5) 
 1 |  6        (1,6)
 2 | -3        (2,-3)
 3 |  2        (3,2)   
 4 | 69        (4,69)   <-- too high to plot.

Then plot them and draw a smooth curve through them:



There is a real zero between where x=-3 and where x=-2
because when x is -3, y is 63, a positive number and
when x=-2, y is -3, a negative number. So for the graph
to get from a y-value that is positive a y-value that is 
negative, the graph must cross the x-axis between 
them. 

There is a real zero between where x=-1 and where x=0
because when x is -1, y is -6, a negative number and
when x=0, y is 5, a negative number. So for the graph
to get from a y-value that is negative to a a y-value 
that is positive, the graph must cross the x-axis between 
them. 

There is a real zero between where x=1 and where x=2
because when x is 1, y is 6, a positive number and
when x=2, y is -3, a negative number. So for the graph
to get from a y-value that is positive a y-value that is 
negative, the graph must cross the x-axis between 
them. 

There is a real zero between where x=2 and where x=3
because when x is 2, y is -3, a negative number and
when x=3, y is 2, a positive number. So for the graph
to get from a y-value that is negative a y-value that is 
positive, the graph must cross the x-axis between 
them. 

Looks like the leftmost relative minimum point ("valley") occurs 
between where x = -2 and -1.

Looks like the only relative maximum point ("peak") occurs 
between where x = 0 and 1.

Looks like the rightmost relative minimum point ("valley") occurs 
between where x = 2 and 3.

Edwin