Question

choose any 5 problems.

1. find any x -intercepts and y -intercepts of f(x)=(x + 3)(x - 2)^2.
2. let h(x)=2(x + 1)^3(x - 5)(x + 2)^2. find the zeros of h and state their multiplicity.
3. the graph of a quintic function g is shown on the xy -coordinate plane. write an equation for g.
4. the graph of a quartic function h is shown on the xy -coordinate plane. write an equation for h.
5. write an equation for a polynomial with x -intercepts at x=-3,x = 1, and x = 4 and a y -intercept -24.

Explanation:

Problem 1

Step1: Find x - intercepts

Set $f(x)=0$, so $(x + 3)(x - 2)^2=0$. By the zero - product property, $x+3 = 0$ or $(x - 2)^2=0$. Solving $x+3=0$ gives $x=-3$, and solving $(x - 2)^2=0$ gives $x = 2$.

Step2: Find y - intercept

Set $x = 0$, then $f(0)=(0 + 3)(0 - 2)^2=3\times4 = 12$.

Step1: Find the zeros

Set $h(x)=0$, so $2(x + 1)^3(x - 5)(x + 2)^2=0$. By the zero - product property, $x+1=0$ or $x - 5=0$ or $x+2=0$. Solving $x+1=0$ gives $x=-1$ with multiplicity 3, solving $x - 5=0$ gives $x = 5$ with multiplicity 1, and solving $x+2=0$ gives $x=-2$ with multiplicity 2.

Step1: Write the polynomial in factored form

Since the x - intercepts are at $x=-3,x = 1,x = 4$, the polynomial has factors $(x + 3),(x - 1),(x - 4)$. So the polynomial is of the form $y=a(x + 3)(x - 1)(x - 4)$.

Step2: Find the value of a

Set $x = 0$ and $y=-24$. Then $-24=a(0 + 3)(0 - 1)(0 - 4)$. Simplify the right - hand side: $a(3)(-1)(-4)=12a$. Solving $12a=-24$ gives $a=-2$.

Step3: Write the final polynomial

Substitute $a=-2$ into the factored form, we get $y=-2(x + 3)(x - 1)(x - 4)$.

Answer:

x - intercepts are $x=-3$ and $x = 2$, y - intercept is $y = 12$

Problem 2