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# Activity 3 - Working with polynomial functions and their graphs Introduction: [Note comments in brackets [?] shoud be pondered; you need not submit a...

Need help understanding Problem 4 on slide 7

Activity 3 – Working with polynomial functions and their graphs Introduction: [ Note comments in brackets [?] shoud be pondered; you need not submit a response.] Consider the function 2 ( ) 6 5 f x x x = - + (1) By inspection, we see this is a quadratic function (or 2 nd degree polynomial function) [why?] and its graph is a parabola that opens upwards. This particular function may be factored relatively nicely and we see that ( ) ( 1)( 5) f x x x = - - (2) Thus, the zeros of the function are x = 1 and x = 5 [why?] If you were to graph this function, the x -intercepts are located at (1, 0) and 5, 0) 6 , or 3 2 2 b x x a - = = = 2 ( ) ( 3) 4 f x x = - - Furthermore, the axis of symmetry is This is also evident when we complete the square and determine The vertex is thus located at the point (3, -4) With this short analysis [and a quick sketch of the function], we can answer many questions:
1. Where is f ( x ) decreasing? [on the interval ] Quick sketch based upon information: 3 , 4 1 , 0 5 , 0 0 2 4 6 1 0 5 0 5 1 0 x y Q u i c k S k e t c h o f y f x ( ,3) - 2. Where is f ( x ) < 0 ? [on the interval ] (1,5) 3 , 4 1 , 0 5 , 0 0 2 4 6 1 0 5 0 5 1 0 x y w h e r e i s f x d e c r e a s i n g ? 3 , 4 1 , 0 5 , 0 0 2 4 6 1 0 5 0 5 1 0 x y w h e r e i s f x   0 ? In general, we say that a local minimum or relative minimum is located at x = 3. We note that the points of interest (where the function changes signs) are the x- intercepts
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Dear Student, Please find attached solution of an... View the full answer

Given polynomial function has x intercept (means y=0) at x=­1,x=2,x­4.
So it should be passing through points (x,y) as(­1,0),(2,0) and (4,0).
Also it should pass through y intercept(x=0) at...

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