The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by: P(x)=-12x^(2)+120 x What is the maximum power generated by the circuit? watts

Identify Parabola Direction: To find the maximum power generated by the circuit, we need to find the vertex of the parabola represented by the quadratic function P(x) = -12x^2 + 120x. Since the coefficient of x^2 is negative, the parabola opens downwards, and the vertex will give us the maximum value of P(x).Calculate Vertex: The vertex of a parabola given by the function f(x) = ax^2 + bx + c is at the point (h, k), where h = -b/(2a). In our case, a = -12 and b = 120.Find x-coordinate of Vertex: Calculate the x-coordinate of the vertex (h) using h = -b/(2a). h = -120 / (2 * -12) h = -120 / -24 h = 5Calculate y-coordinate of Vertex: Now we need to calculate the y-coordinate of the vertex (k), which is the maximum power P(h). We do this by substituting x = h into the function P(x). P(h) = -12h^2 + 120hSubstitute x into Function: Substitute h = 5 into P(h) to find the maximum power. P(5) = -12(5)^2 + 120(5) P(5) = -12(25) + 600 P(5) = -300 + 600 P(5) = 300Calculate Maximum Power: The maximum power generated by the circuit is 300 watts.

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The power generated by an electrical circuit (in watts) as a function of its current
x (in amperes) is modeled by:

P(x)=-12x^(2)+120 x
What is the maximum power generated by the circuit?
watts

The power generated by an electrical circuit (in watts) as a function of its current $x$ (in amperes) is modeled by: $\newline$ $P(x)=-12x^{2}+120x$ $\newline$ What is the maximum power generated by the circuit? $\newline$ $\text{watts}$

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Q. The power generated by an electrical circuit (in watts) as a function of its current

x

(in amperes) is modeled by:

\newline

P(x)=-12x^{2}+120x

\newline

What is the maximum power generated by the circuit?

\newline

\text{watts}

Identify Parabola Direction: To find the maximum power generated by the circuit, we need to find the vertex of the parabola represented by the quadratic function $P(x) = -12x^2 + 120x$ . Since the coefficient of $x^2$ is negative, the parabola opens downwards, and the vertex will give us the maximum value of $P(x)$ .
Calculate Vertex: The vertex of a parabola given by the function $f(x) = ax^2 + bx + c$ is at the point $(h, k)$ , where $h = -\frac{b}{2a}$ . In our case, $a = -12$ and $b = 120$ .
Find x-coordinate of Vertex: Calculate the x-coordinate of the vertex $h$ using $h = -\frac{b}{2a}$ . $h = -\frac{120}{2 \times -12}$ $h = -\frac{120}{-24}$ $h = 5$
Calculate y-coordinate of Vertex: Now we need to calculate the y-coordinate of the vertex $k$ , which is the maximum power $P(h)$ . We do this by substituting $x = h$ into the function $P(x)$ . $P(h) = -12h^2 + 120h$
Substitute $x$ into Function: Substitute $h = 5$ into $P(h)$ to find the maximum power. $\newline$ $P(5) = -12(5)^2 + 120(5)$ $\newline$ $P(5) = -12(25) + 600$ $\newline$ $P(5) = -300 + 600$ $\newline$ $P(5) = 300$
Calculate Maximum Power: The maximum power generated by the circuit is $300$ watts.