Question

1. sketch an even degree function with these characteristics: ( a > 0 ) and it has 2 distinct real zeros and 2 imaginary zeros.

Explanation:

Step1: Determine the degree

An even - degree function with 2 distinct real zeros and 2 imaginary zeros. Since imaginary zeros come in conjugate pairs, and we have 2 imaginary zeros (a pair) and 2 real zeros. The total number of zeros (counting multiplicities? No, distinct here) but for the degree, the degree is equal to the sum of the number of real and non - real (imaginary) zeros (since non - real zeros come in pairs). So the degree should be \(2 + 2=4\) (even degree, which matches the requirement).

Step2: Form the function

For a polynomial function \(f(x)=a(x - r_1)(x - r_2)(x-(m + ni))(x-(m - ni))\), where \(a>0\), \(r_1\) and \(r_2\) are real zeros, and \(m+ni\) and \(m - ni\) are imaginary zeros. Let's choose simple real zeros, say \(r_1 = 1\), \(r_2=2\) and imaginary zeros \(m + ni=i\), \(m - ni=-i\) (so \(m = 0\), \(n = 1\)). Then the function is \(f(x)=a(x - 1)(x - 2)(x - i)(x + i)\). Since \((x - i)(x + i)=x^{2}+1\) and \((x - 1)(x - 2)=x^{2}-3x + 2\), then \(f(x)=a(x^{2}-3x + 2)(x^{2}+1)\). Let's take \(a = 1\) (since \(a>0\)), so \(f(x)=(x^{2}-3x + 2)(x^{2}+1)=x^{4}-3x^{3}+3x^{2}-3x + 2\).

Step3: Analyze the end - behavior

For a polynomial \(f(x)=a_nx^n+\cdots+a_0\) with \(n = 4\) (even) and \(a_n=a = 1>0\), as \(x
ightarrow\pm\infty\), \(f(x)
ightarrow\infty\) (because the leading term \(x^{4}\) dominates and for large \(|x|\), \(x^{4}\) is positive and large).

Step4: Analyze the real zeros

The real zeros are at \(x = 1\) and \(x = 2\) (from \((x - 1)(x - 2)\)).

Step5: Sketch the graph

• The graph will cross the x - axis at \(x = 1\) and \(x = 2\) (since the multiplicity of each real zero is 1, which is odd, so the graph crosses the x - axis at these points).
• The end - behavior is that as \(x

ightarrow\infty\) and \(x
ightarrow-\infty\), the graph goes up (because the leading coefficient is positive and degree is even).

• Between the real zeros (1 and 2), we can test a value, say \(x = 1.5\). \(f(1.5)=( (1.5)^{2}-3\times1.5 + 2)((1.5)^{2}+1)=(2.25-4.5 + 2)(2.25 + 1)=( - 0.25)(3.25)=- 0.8125<0\). So the graph dips below the x - axis between 1 and 2.
• For \(x<1\), test \(x = 0\): \(f(0)=(0 - 0+2)(0 + 1)=2>0\). For \(x>2\), test \(x = 3\): \(f(3)=(9 - 9+2)(9 + 1)=(2)(10)=20>0\).

Answer:

A possible function is \(f(x)=(x - 1)(x - 2)(x^{2}+1)=x^{4}-3x^{3}+3x^{2}-3x + 2\) (with \(a = 1>0\)) and its graph has end - behavior going to \(+\infty\) as \(x
ightarrow\pm\infty\), crosses the x - axis at \(x = 1\) and \(x = 2\), and dips below the x - axis between \(x = 1\) and \(x = 2\). The sketch should reflect these characteristics.