Question

construct & justify
create a polynomial function that meets all of the following criteria:

• it must be degree-3 or degree-4.
• it must have exactly two turning points.
• it must have a specific y-intercept (you choose the value).

your response must include three parts:

1. the function itself, written in standard form.
2. a brief description of a sketch of the graph.
3. 2-3 sentences explaining how the degree and leading coefficient of your function create the end behavior, and how you ensured there were exactly two turning points.

Explanation:

Step1: Choose Degree and Form

A degree - 3 polynomial has at most 2 turning points (since the number of turning points of a polynomial of degree \(n\) is at most \(n - 1\)). Let's choose a cubic polynomial. Let the y - intercept be 5 (so when \(x = 0\), \(y=5\)). We can start with a factored form and expand it. Let's consider a polynomial with a double root and a single root. Let the polynomial be \(f(x)=(x - 1)^2(x + 1)\).
First, expand \((x - 1)^2\): \((x - 1)^2=x^{2}-2x + 1\).
Then multiply by \((x + 1)\): \((x^{2}-2x + 1)(x + 1)=x^{3}+x^{2}-2x^{2}-2x+x + 1=x^{3}-x^{2}-x + 1\). But we want a y - intercept of 5. So we can adjust the polynomial. Let's consider \(f(x)=x^{3}-x^{2}-x + 5\).

Step2: Analyze Turning Points

To find the number of turning points, we take the derivative \(f^\prime(x)=3x^{2}-2x - 1\). The number of real roots of the derivative will give the number of turning points. The discriminant of \(ax^{2}+bx + c\) is \(\Delta=b^{2}-4ac\). For \(3x^{2}-2x - 1\), \(a = 3\), \(b=-2\), \(c=-1\), \(\Delta=(-2)^{2}-4\times3\times(-1)=4 + 12 = 16>0\), so the derivative has two real roots, meaning the function has two turning points.

Step3: Analyze End Behavior

For a cubic polynomial \(f(x)=ax^{3}+bx^{2}+cx + d\), the end behavior is determined by the leading term \(ax^{3}\). Here \(a = 1>0\), so as \(x
ightarrow\infty\), \(f(x)
ightarrow\infty\) and as \(x
ightarrow-\infty\), \(f(x)
ightarrow-\infty\). The degree is 3 (odd) and leading coefficient positive, so the graph goes down to the left and up to the right. The y - intercept is 5 (when \(x = 0\), \(y = 5\)). The graph will have a local maximum and a local minimum (two turning points) because the derivative has two real critical points.

Answer:

1. The polynomial function in standard form: \(f(x)=x^{3}-x^{2}-x + 5\)
2. Sketch description: The graph is a cubic curve. It crosses the y - axis at \((0,5)\). As \(x

ightarrow-\infty\), the graph goes down, and as \(x
ightarrow\infty\), it goes up. It has a local maximum and a local minimum (two turning points) between the regions where the function is increasing and decreasing.

1. Explanation: The function is degree - 3 (odd) with a positive leading coefficient (\(a = 1\)), so as \(x

ightarrow\infty\), \(f(x)
ightarrow\infty\) and as \(x
ightarrow-\infty\), \(f(x)
ightarrow-\infty\). The derivative \(f^\prime(x)=3x^{2}-2x - 1\) has two real roots (since its discriminant \(\Delta = 16>0\)), so the function has exactly two turning points.