$\begin{array}{l}y=4-x^{2} \\ y=3 x\end{array}$

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Find higher derivatives of rational and radical functions

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\begin{array}{l}y=4-x^{2} \\ y=3 x\end{array}

Set Equations Equal: To find the points of intersection between the parabola $y=4-x^2$ and the line $y=3x$ , we need to set the two equations equal to each other and solve for $x$ . So, we set $4 - x^2 = 3x$ .
Rearrange and Form Quadratic Equation: Rearrange the equation to bring all terms to one side, which will give us a quadratic equation. $x^2 + 3x - 4 = 0$
Factor the Quadratic Equation: Now we need to solve the quadratic equation. We can do this by factoring, completing the square, or using the quadratic formula. The equation looks like it can be factored easily. $\newline$ Let's try to factor the quadratic equation.
Find Values of $x$ : We look for two numbers that multiply to $-4$ and add up to $3$ . Those numbers are $4$ and $-1$ . So, we can write the equation as $(x + 4)(x - 1) = 0$ .
Solve for x: Now we can find the values of $x$ by setting each factor equal to zero. $x + 4 = 0$ or $x - 1 = 0$
Substitute $x$ into Original Equation: Solving for $x$ gives us two solutions: $x = -4$ or $x = 1$
Identify Points of Intersection: We now substitute these $x$ -values back into either of the original equations to find the corresponding $y$ -values. Let's use $y = 3x$ since it's simpler. $\newline$ For $x = -4$ : $y = 3(-4) = -12$ $\newline$ For $x = 1$ : $y = 3(1) = 3$
Identify Points of Intersection: We now substitute these $x$ -values back into either of the original equations to find the corresponding $y$ -values. Let's use $y = 3x$ since it's simpler. $\newline$ For $x = -4$ : $y = 3(-4) = -12$ $\newline$ For $x = 1$ : $y = 3(1) = 3$ We have found two points of intersection: $(-4, -12)$ and $(1, 3)$ .