
To sketch: the graph of the given polynomial function.

Explanation of Solution
Given information:
A polynomial function is given as
h(x)=x4−2x3+3x
Concept used:
A polynomial function is of the form
f(x)=anxn+an−1xn−1+...+a1x+a0
Terms of a polynomial function should be arranged in descending order according to its degree to express it in a standard form and degree of each term should be a positive integer or whole number. The coefficients should be real numbers.
Leading coefficient of a polynomial function is the coefficient of the leading term.
Degree of the polynomial is the degree of leading term or the height degree in the polynomial function.
For the polynomial f(x)=anxn+an−1xn−1+...+a1x+a0 the leading coefficient is an and the degree is n .
The end behavior can describe the graph of a polynomial function as x approaches +∞ or as x approaches −∞ .
The end behavior of a polynomial function can be determined by the leading coefficient and the degree of the polynomial.
If degree is even and leading coefficient is negative.
f(x)→−∞ as x→−∞ and f(x)→−∞ as x→+∞ .
If degree is odd and leading coefficient is negative.
f(x)→∞ as x→−∞ and f(x)→−∞ as x→+∞ .
If degree is odd and leading coefficient is positive.
f(x)→−∞ as x→−∞ and f(x)→∞ as x→+∞ .
If degree is even and leading coefficient is positive.
f(x)→∞ as x→−∞ and f(x)→∞ as x→+∞ .
Calculation:
Consider the given polynomial function.
h(x)=x4−2x3+3x
Now, tabulate some points as shown below:
x | h(x)=x4−2x3+3x | (x,h(x)) |
0 | h(0)=(0)4−2(0)3+3(0)=0 | (0,0) |
1 | h(1)=(1)4−2(1)3+3(1)=2 | (1,2) |
−1 | h(−1)=(1)4−2(−1)3+3(1)=6 | (−1,6) |
2 | h(2)=(2)4−2(2)3+3(2)=6 | (2,6) |
−2 | h(−2)=(−2)4−2(−2)3+3(−2)=26 | (−2,26) |
Graph:
Plot the points and draw a smooth curve to sketch the graph.
Interpretation:
Degree is even and leading coefficient is positive.
h(x)→∞ as x→−∞ and h(x)→∞ as x→+∞ .
Chapter 4 Solutions
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