Question

For this problem, we will consider the polynomial function f(z) = 4x4 - 1623 +2422 - 322 +32 over the interval -3 <I<3 (a) The degree of f(x) is Number (b) Which of the following choices describe the end behavior of f(x)? The graph of f(x) acts O like 22 (ie both ends up) Olike -3 (ie both ends down) like 23 (ie left end down, right end up) Olike -23 -(e. left end up, right end down) O None of the above (c) Following the examples from the book (and video) use a synthetic division chart to generate a list of points to plot Type these points as a comma-separated list where each point is entered using the form (x, y) (d) Sketch a graph of f(x) = 424 - 1623 + 24 22 - 32 x + 32 (on a piece of scrap paper) and then choose the correct graph below.

Answer 1

a)Given is a polynomial fucntion in one variable i.e. x . As per the rule, the degree is the highest power of the variable in the fucntion/expression. Lets consider an example : x²+4x+16. The degree of the equation is 2 .i.e. the highest power of variable.

Thus the degree of f(x) is : 4

b)The leading term of the polynomial with highest power is 4x⁴. Thus we can see from this term that the highest degree of given polynomial is even and the leading coefficient of this term is positive (4).

This states that the when x approaches infinity(∞) , f(x) also approaches infinity(∞ ) and

when x approaches negative infinity (-∞), f(x) again approaches infinity(∞ ).

Hence the graph of f(x) acts like x² (i.e. both ends up)   therefore the first option is correct and hence the graph cannot behave like x to the power -2 or 3 or -3. hence other options are wrong.

c)One of the point is [2, 0 ]. the working is shown in the image below.

Date c) Working Equation Dividend): 4x4 - 1623 +2422-322 +28 (ie (x-2) is the root) Lets try by This synthetic division x' -16 C 32 24 - 32 니 2 a 4 - 32 32 24) 8 8 그 - 16+ 8 = 8 Adding (3) Similarly a u 16 -32 24 - 16 32 - 32 8 3 '4 Thus the quatient is and 8 x ² + 8x -16 remainder Therefore y = f(x) = (x-2) (4x3 - 83c² +8x-16) put (x = aj we get 4= f(): () (u(2)° - 8 (a) * 48(2)-16) Ty=o Thus (x, y] =[2,o] to be plotted graph Similarly other points can be found out a point on

d)The solution is given in image below

d) f(x) f(x) = 404_1603 +24x² – 32x+32 Ef(x) if(x) g-axio o Page No.: 5 3 DC axis 2 3

e)There is only one x-intercept : 2

The x- intercepts can be found by putting y= 0 , in the given equation.(working shown in image)

e) This y -1603 +2402 -32x +32 10=4x 4-16x² + 24x2 32 oC +32 4004 ) Factor the 4 (x-2) (x-2) (x²+2) right hand side - Q -2) = 0 x² +2=0 c=2 2. -2 x = 2 the ra are 2 Thus Therefore real roots Ic-intercept = 2

f)the graph has one(1) turning point at x = 2. As seen from the image of the graph in part d.

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