c In any function, the constant tells us where it intercepts the y-axis. In this case, it would appear that there is no constant on the right-hand side. But when a constant is missing, it implies that it is 0.
q(x)=x^2+5x+ 0 ← constant
Therefore, the y-intercept is at the origin, (0,0). To find the x-intercept(s), we will as in Part B, set the function equal to 0 and solve for x.
The x-intercepts are (- 5,0) and (0,0). Next we want to draw the graphs of p(x) and q(x). Let's first mark the x- and y-intercepts for each function in a diagram.
To graph this accurately, we need to find some more points for each graph using a value table.
|c|c|c|
[-1em]
x & x^2+5x-6 & p(x) [0.2em]
[-1em]
-5 & ( -5)^2+5( -5)-6 & -6 [0.2em]
[-1em]
-4 & ( -4)^2+5( -4)-6 & -10 [0.2em]
[-1em]
-3 & ( -3)^2+5( -3)-6 & -12 [0.2em]
[-1em]
-2 & ( -2)^2+5( -2)-6 & -12 [0.2em]
[-1em]
-1 & ( -1)^2+5( -1)-6 & -10 [0.2em]
Let's also do this for g(x). We also want to know the corresponding points on q(x). Let's make a table of values for this function.
|c|c|c|
[-1em]
x & x^2+5x & q(x) [0.2em]
[-1em]
-4 & ( -4)^2+5( -4) & -4 [0.2em]
[-1em]
-3 & ( -3)^2+5( -3) & -6 [0.2em]
[-1em]
-2 & ( -2)^2+5( -2) & -6 [0.2em]
[-1em]
-1 & ( -1)^2+5( -1) & -4 [0.2em]
Let's mark the points in our coordinate plane and connect each set of points with a smooth curve.
Examining the diagram, we notice that p(x) is a vertical translation of q(x) in the negative direction. To determine how much it has translated, we must find the vertical distance between two corresponding points.
As we can see, the graph of p(x) is 6 units below q(x).
d By substituting q(x) and p(x) with the right-hand sides of the functions, we can calculate the difference by combining like terms.
