Question

1. determine an equation for the family of quartic functions with zeros 3 (order 2) and -2 ± √5: b) determine the equation for the member of the family whose graph passes through the point (-1,4) c) sketch the graph in (b)

Explanation:

Step1: Find the general equation of the quartic function

A quartic function (degree 4) with zeros \( r_1, r_2, \dots, r_n \) (with multiplicities) can be written in factored form as \( f(x) = a(x - r_1)^{m_1}(x - r_2)^{m_2}\dots(x - r_n)^{m_n} \), where \( a \) is a non - zero constant and \( m_i \) are the multiplicities (orders) of the zeros.

The zero \( 3 \) has order \( 2 \), so the factor corresponding to it is \( (x - 3)^2 \).

The zeros \( - 2+\sqrt{5} \) and \( - 2-\sqrt{5} \) are roots of the quadratic equation \( (x + 2)^2-5=0 \) (since if \( x=-2\pm\sqrt{5} \), then \( (x + 2)^2=5 \), or \( x^{2}+4x + 4 - 5=x^{2}+4x - 1 = 0 \)). The factors corresponding to these zeros are \( (x-(-2 + \sqrt{5}))(x-(-2-\sqrt{5}))=(x + 2-\sqrt{5})(x + 2+\sqrt{5})=(x + 2)^2-(\sqrt{5})^2=x^{2}+4x+4 - 5=x^{2}+4x - 1 \).

So the general equation of the quartic function is \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \), where \( a eq0 \).

Step2: Substitute the point \((-1,4)\) into the equation to find \( a \)

We know that the graph of the function passes through the point \((-1,4)\). Substitute \( x=-1 \) and \( f(-1) = 4 \) into the equation \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \):

First, calculate \( (x - 3)^2 \) when \( x=-1 \): \( (-1 - 3)^2=(-4)^2 = 16 \)

Then, calculate \( x^{2}+4x - 1 \) when \( x=-1 \): \( (-1)^2+4\times(-1)-1=1-4 - 1=-4 \)

Now, substitute these values into the function: \( 4=a\times16\times(-4) \)

Simplify the right - hand side: \( 4=-64a \)

Solve for \( a \): \( a=\frac{4}{-64}=-\frac{1}{16} \)

Step3: Write the equation of the function

Substitute \( a =-\frac{1}{16} \) into the general form \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \):

\( f(x)=-\frac{1}{16}(x - 3)^2(x^{2}+4x - 1) \)

We can also expand this function:

First, expand \( (x - 3)^2=x^{2}-6x + 9 \)

Then, multiply \( (x^{2}-6x + 9)(x^{2}+4x - 1) \)

\[ \begin{align*} &x^{2}(x^{2}+4x - 1)-6x(x^{2}+4x - 1)+9(x^{2}+4x - 1)\\ =&x^{4}+4x^{3}-x^{2}-6x^{3}-24x^{2}+6x + 9x^{2}+36x - 9\\ =&x^{4}+(4x^{3}-6x^{3})+(-x^{2}-24x^{2}+9x^{2})+(6x + 36x)-9\\ =&x^{4}-2x^{3}-16x^{2}+42x - 9 \end{align*} \]

Then \( f(x)=-\frac{1}{16}(x^{4}-2x^{3}-16x^{2}+42x - 9)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \)

Step4: Sketch the graph (for part c)

1. **End - behavior**: The leading term of the quartic function \( f(x)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \) is \( y =-\frac{1}{16}x^{4} \). Since the leading coefficient \( a=-\frac{1}{16}<0 \) and the degree \( n = 4 \) (even), as \( x ightarrow\pm\infty \), \( y ightarrow-\infty \).
2. **Zeros**:
- The zero \( x = 3 \) has multiplicity 2, so the graph touches the \( x \) - axis at \( x = 3 \) and turns around.
- The zeros \( x=-2\pm\sqrt{5}\approx-2\pm2.24 \), so \( x\approx0.24 \) and \( x\approx - 4.24 \). The graph crosses the \( x \) - axis at these points.
3. **Y - intercept**: To find the \( y \) - intercept, set \( x = 0 \). \( f(0)=-\frac{1}{16}(0 - 3)^2(0^{2}+4\times0 - 1)=-\frac{1}{16}\times9\times(-1)=\frac{9}{16}=0.5625 \)
4. **Point plotting**: We know the point \((-1,4)\) is on the graph. We can also find a few more points by plugging in some \( x \) - values (e.g., \( x = 1 \): \( f(1)=-\frac{1}{16}(1 - 3)^2(1^{2}+4\times1 - 1)=-\frac{1}{16}\times4\times4=-1 \); \( x = 2 \): \( f(2)=-\frac{1}{16}(2 - 3)^2(2^{2}+4\times2 - 1)=-\frac{1}{16}\times1\times11=-\frac{11}{16}\approx - 0.69 \))

Using the end - behavior, zeros, \( y \) - intercept, and additional points, we can sketch the graph. The graph will have a "W" - like or "M" - like shape (but since the leading coefficient is negative, it will open downwards at the ends) with a touch point at \( x = 3 \) and cross points at \( x=-2\pm\sqrt{5} \), and pass through \((-1,4)\) and \((0,\frac{9}{16})\) etc.

Part (a) Answer: The family of quartic functions is \( f(x)=a(x - 3)^2(x^{2}+4x - 1),a

eq0 \)

Part (b) Answer: The equation of the function is \( f(x)=-\frac{1}{16}(x - 3)^2(x^{2}+4x - 1) \) (or the expanded form \( f(x)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \))

Part (c) Answer: The graph is sketched using the end - behavior, zeros, \( y \) - intercept, and key points as described above.

Answer:

Step1: Find the general equation of the quartic function

A quartic function (degree 4) with zeros \( r_1, r_2, \dots, r_n \) (with multiplicities) can be written in factored form as \( f(x) = a(x - r_1)^{m_1}(x - r_2)^{m_2}\dots(x - r_n)^{m_n} \), where \( a \) is a non - zero constant and \( m_i \) are the multiplicities (orders) of the zeros.

The zero \( 3 \) has order \( 2 \), so the factor corresponding to it is \( (x - 3)^2 \).

The zeros \( - 2+\sqrt{5} \) and \( - 2-\sqrt{5} \) are roots of the quadratic equation \( (x + 2)^2-5=0 \) (since if \( x=-2\pm\sqrt{5} \), then \( (x + 2)^2=5 \), or \( x^{2}+4x + 4 - 5=x^{2}+4x - 1 = 0 \)). The factors corresponding to these zeros are \( (x-(-2 + \sqrt{5}))(x-(-2-\sqrt{5}))=(x + 2-\sqrt{5})(x + 2+\sqrt{5})=(x + 2)^2-(\sqrt{5})^2=x^{2}+4x+4 - 5=x^{2}+4x - 1 \).

So the general equation of the quartic function is \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \), where \( a
eq0 \).

Step2: Substitute the point \((-1,4)\) into the equation to find \( a \)

We know that the graph of the function passes through the point \((-1,4)\). Substitute \( x=-1 \) and \( f(-1) = 4 \) into the equation \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \):

First, calculate \( (x - 3)^2 \) when \( x=-1 \): \( (-1 - 3)^2=(-4)^2 = 16 \)

Then, calculate \( x^{2}+4x - 1 \) when \( x=-1 \): \( (-1)^2+4\times(-1)-1=1-4 - 1=-4 \)

Now, substitute these values into the function: \( 4=a\times16\times(-4) \)

Simplify the right - hand side: \( 4=-64a \)

Solve for \( a \): \( a=\frac{4}{-64}=-\frac{1}{16} \)

Step3: Write the equation of the function

Substitute \( a =-\frac{1}{16} \) into the general form \( f(x)=a(x - 3)^2(x^{2}+4x - 1) \):

\( f(x)=-\frac{1}{16}(x - 3)^2(x^{2}+4x - 1) \)

We can also expand this function:

First, expand \( (x - 3)^2=x^{2}-6x + 9 \)

Then, multiply \( (x^{2}-6x + 9)(x^{2}+4x - 1) \)

$$
LATEXBLOCK0
$$

Then \( f(x)=-\frac{1}{16}(x^{4}-2x^{3}-16x^{2}+42x - 9)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \)

Step4: Sketch the graph (for part c)

1. End - behavior: The leading term of the quartic function \( f(x)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \) is \( y =-\frac{1}{16}x^{4} \). Since the leading coefficient \( a=-\frac{1}{16}<0 \) and the degree \( n = 4 \) (even), as \( x

ightarrow\pm\infty \), \( y
ightarrow-\infty \).

1. Zeros:

• The zero \( x = 3 \) has multiplicity 2, so the graph touches the \( x \) - axis at \( x = 3 \) and turns around.
• The zeros \( x=-2\pm\sqrt{5}\approx-2\pm2.24 \), so \( x\approx0.24 \) and \( x\approx - 4.24 \). The graph crosses the \( x \) - axis at these points.

1. Y - intercept: To find the \( y \) - intercept, set \( x = 0 \). \( f(0)=-\frac{1}{16}(0 - 3)^2(0^{2}+4\times0 - 1)=-\frac{1}{16}\times9\times(-1)=\frac{9}{16}=0.5625 \)
2. Point plotting: We know the point \((-1,4)\) is on the graph. We can also find a few more points by plugging in some \( x \) - values (e.g., \( x = 1 \): \( f(1)=-\frac{1}{16}(1 - 3)^2(1^{2}+4\times1 - 1)=-\frac{1}{16}\times4\times4=-1 \); \( x = 2 \): \( f(2)=-\frac{1}{16}(2 - 3)^2(2^{2}+4\times2 - 1)=-\frac{1}{16}\times1\times11=-\frac{11}{16}\approx - 0.69 \))

Using the end - behavior, zeros, \( y \) - intercept, and additional points, we can sketch the graph. The graph will have a "W" - like or "M" - like shape (but since the leading coefficient is negative, it will open downwards at the ends) with a touch point at \( x = 3 \) and cross points at \( x=-2\pm\sqrt{5} \), and pass through \((-1,4)\) and \((0,\frac{9}{16})\) etc.

Part (a) Answer: The family of quartic functions is \( f(x)=a(x - 3)^2(x^{2}+4x - 1),a

eq0 \)

Part (b) Answer: The equation of the function is \( f(x)=-\frac{1}{16}(x - 3)^2(x^{2}+4x - 1) \) (or the expanded form \( f(x)=-\frac{1}{16}x^{4}+\frac{1}{8}x^{3}+x^{2}-\frac{21}{8}x+\frac{9}{16} \))

Part (c) Answer: The graph is sketched using the end - behavior, zeros, \( y \) - intercept, and key points as described above.