
Concept explainers
To find: The standard form of the quadratic function and the graph for it, also identify the graph, vertex axis of symmetry and x intercepts.

Answer to Problem 26E
The quadratic function in the standard form is f(x)=x2−30x+225 , the vertex of the graph is (15,0) and its axis of symmetry is x=15 , the x intercepts is (15,0) and the graph is shown in Figure 1
Explanation of Solution
Given:
The given function is f(x)=x2−30x+225 .
Calculation:
Consider the quadratic function is,
f(x)=a(x−h)2+k
The vertex of the above parabola for the standard form f(x)=a(x−h)2+k is,
(h,k)
Consider the given equation is,
f(x)=x2−30x+225=(x−15)2
From the above equation the vertex of the function f(x)=x2−30x+225 is,
(h,k)=(15,0)
Thus, the axis of symmetry of the graph is x=15 .
To find the x intercept of the function x2−30x+225=0 equate the expression equal to zero as,
x2−30x+225=0(x−15)2=0x=15
There is only one intercept of the x axis this suggests that the parabola does not cut the x axis but touches it at the point (15,0)
Consider the point for x=11 then the function,
f(11)=112−330+225=16
Consider the point for x=19 , then the function is,
f(19)=361−570+225=16
The sign of the coefficient a is positive so the graph for the parabola will open in the upward direction with the vertices at (15,0) across the point (11,76), (19,16) .
The graph is shown in Figure 1
Figure 1
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