Question

question 3: y - intercept: - 10 zeros and multiplicity: - 2, 5 2, 1 end behavior: $lim_{x
ightarrowinfty}g(x)=infty$, $lim_{x
ightarrow-infty}g(x)=-infty$ degree: 3 leading coefficient: equation for $g(x)$ in intercept form: challenge! equation for $g(x)$ in standard form: question 4: y - intercept: zeros and multiplicity: end behavior: $lim_{x
ightarrowinfty}h(x)=$, $lim_{x
ightarrow-infty}h(x)=$ degree: 4 leading coefficient: equation for $h(x)$ in intercept form: challenge! equation for $h(x)$ in standard form:

Explanation:

Question 3 - Function $g(x)$

Step1: Identify y - intercept

The y - intercept is the value of the function when $x = 0$. From the graph, when $x=0$, $g(0)=- 10$.

Step2: Identify zeros and multiplicity

The zeros of the function are the x - values where $g(x)=0$. From the graph, the zeros are $x=-2$ and $x = 5$. If the graph crosses the x - axis at a zero, the multiplicity is odd. Here, the multiplicity of $x=-2$ is 1 and of $x = 5$ is 1.

Step3: Determine end - behavior

As $x\to-\infty$, $g(x)\to-\infty$ and as $x\to+\infty$, $g(x)\to+\infty$. Since the end - behavior is opposite, the degree of the polynomial is odd. Given the number of zeros and the shape, the degree is 3. For a cubic function $y = ax^{3}+bx^{2}+cx + d$ with end - behavior $x\to-\infty,y\to-\infty$ and $x\to+\infty,y\to+\infty$, the leading coefficient $a>0$.
The intercept form of a polynomial is $y=a(x - r_1)(x - r_2)(x - r_3)$. Using the zeros $r_1=-2,r_2 = 5$ and the y - intercept $y(0)=-10$, we substitute $x = 0$ and $y=-10$ into $y=a(x + 2)(x - 5)$.
\[

$$\begin{align*} -10&=a(0 + 2)(0 - 5)\\ -10&=a(2)(-5)\\ -10&=-10a\\ a&=1 \end{align*}$$

\]
So the intercept form is $g(x)=(x + 2)(x - 5)(x - k)$ (we can assume $k$ is a non - real root or a root with multiplicity 0 in the context of the real - valued graph). Expanding $(x + 2)(x - 5)=x^{2}-3x - 10$, then multiplying by $x$ (since degree is 3) gives the standard form $g(x)=x^{3}-3x^{2}-10x$.

Question 4 - Function $h(x)$

Step1: Identify y - intercept

The y - intercept is the value of the function when $x = 0$. From the graph, when $x = 0$, $h(0)=2$.

Step2: Identify zeros and multiplicity

The graph touches the x - axis at $x=-2$ and crosses at $x = 2$. The zero $x=-2$ has an even multiplicity (since the graph touches the x - axis), say 2, and $x = 2$ has a multiplicity of 1. The degree of the polynomial is the sum of the multiplicities, so the degree is 3.

Step3: Determine end - behavior

As $x\to+\infty$, the graph of $h(x)\to-\infty$ and as $x\to-\infty$, $h(x)\to+\infty$. So the leading coefficient is negative.
The intercept form of the polynomial is $h(x)=a(x + 2)^{2}(x - 2)$. Using the y - intercept $h(0)=2$, we substitute $x = 0$ and $y = 2$ into $h(x)=a(x + 2)^{2}(x - 2)$.
\[

$$\begin{align*} 2&=a(0+2)^{2}(0 - 2)\\ 2&=a(4)(-2)\\ 2&=-8a\\ a&=-\frac{1}{4} \end{align*}$$

\]
The intercept form is $h(x)=-\frac{1}{4}(x + 2)^{2}(x - 2)$. Expanding:
\[

$$\begin{align*} h(x)&=-\frac{1}{4}(x^{2}+4x + 4)(x - 2)\\ &=-\frac{1}{4}(x^{3}-2x^{2}+4x^{2}-8x+4x - 8)\\ &=-\frac{1}{4}(x^{3}+2x^{2}-4x - 8)\\ &=-\frac{1}{4}x^{3}-\frac{1}{2}x^{2}+x + 2 \end{align*}$$

\]

Answer:

Question 3 - $g(x)$
Y - intercept: $-10$
Zeros and multiplicity: $x=-2$ (multiplicity 1), $x = 5$ (multiplicity 1)
End - behavior: $\lim_{x\to-\infty}g(x)=-\infty$, $\lim_{x\to+\infty}g(x)=+\infty$
Degree: 3
Leading coefficient: 1
Intercept form: $g(x)=(x + 2)(x - 5)x$
Standard form: $g(x)=x^{3}-3x^{2}-10x$

Question 4 - $h(x)$
Y - intercept: 2
Zeros and multiplicity: $x=-2$ (multiplicity 2), $x = 2$ (multiplicity 1)
End - behavior: $\lim_{x\to-\infty}h(x)=+\infty$, $\lim_{x\to+\infty}h(x)=-\infty$
Degree: 3
Leading coefficient: $-\frac{1}{4}$
Intercept form: $h(x)=-\frac{1}{4}(x + 2)^{2}(x - 2)$
Standard form: $h(x)=-\frac{1}{4}x^{3}-\frac{1}{2}x^{2}+x + 2$