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The equation
s=(t+3)^(2)(t+2)(t+1)(t)(t-1) is graphed on the
st-plane. What is the product of the unique
t-intercepts of the graph?

The equation $s=(t+3)^{2}(t+2)(t+1)(t)(t-1)$ is graphed on the $s t$ -plane. What is the product of the unique $t$ -intercepts of the graph?

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Find the vertex of the transformed function

Full solution

Q. The equation

s=(t+3)^{2}(t+2)(t+1)(t)(t-1)

is graphed on the

s t

-plane. What is the product of the unique

t

-intercepts of the graph?

Set Equation to Zero: The $t$ -intercepts of a graph occur where the function $s$ equals zero. To find the $t$ -intercepts, we need to set the equation $s$ equal to zero and solve for $t$ . $s = (t+3)^2(t+2)(t+1)(t)(t-1) = 0$
Factor and Solve: Since the equation is already factored, we can find the t-intercepts by setting each factor equal to zero. $\newline$ $(t+3)^2 = 0$ , $(t+2) = 0$ , $(t+1) = 0$ , $t = 0$ , $(t-1) = 0$
Identify T-Intercepts: Solving each equation for $t$ gives us the t-intercepts: $t = -3$ , $t = -2$ , $t = -1$ , $t = 0$ , $t = 1$ Note that $t = -3$ is a repeated root because of the squared term $(t+3)^2$ .
Calculate Product: The product of the unique t-intercepts is found by multiplying the distinct values of $t$ together. $\newline$ Product = $(-3) \times (-2) \times (-1) \times (0) \times (1)$
Final Result: Multiplying the numbers together, we find the product: $\newline$ Product = $0$ $\newline$ Since any number multiplied by zero is zero, the product of the $t$ -intercepts is $0$ .

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